
Class T S 1 K ^ \ ' 

Rnok Y-^ 

Copyright }1^ 



COPYRIGHT DEPOSIT. 



A PRACTICAL TREATISE 



YARN AND CLOTH CALCULATIONS 



COTTON FABRICS 



THOMAS YATES 

Instructor in W^eaving, IVarp Preparation and Calculations 
New Bedford Textile School. 



NEW BEDFORD, MASS. 
1904. 



-^-^ 



LIBRARY of 0ON6RESS 
TVmOopiesRwetved 
SEP 7 1904 



«tAS8 CL xxe. Na 
OePYB 



Entered according to Act of Congress, in the year 1904, 

by Thomas Yates, 

In the oflSce of the Librarian of Congress, at Washington, D. C. 






Preface to Second Edition. 



The demand for the first edition of ''A Practml Treatise 
on Yarn and Cloth Calculations for Cotton Fabrics'' has been 
such that the author has been encouraged to present a 
second edition. 

This second edition has been enlarged by the addition oi 
new calculations and explanations of the rules given for 
their solution. That this volume in its new form may meet 
with the same approval as the first edition is the hope of 
the author. 

TEOMAS YATES. 

New Bedford^ 1904. 



YARN CALCULATIONS. 



In dealing with Yarn Calculations it is necessary to 
have two tal>les or standards, one of leno^th and one of 
weight. 

The standards of length for different materials vary, 
l)ut the calculations are all based on a given length that 
is contained in a given weight. 

Tables of lengths and weights will be given, with 
rules for calculations under the specified materials. 

The table of lengths for Cotton Yarn is as follows : 

11. yds. = 1 thread or cir. of wrap reel, 
120 " = 80 threads or 1 skein or lea, 
840 " — 560 " =7 skeins " " =: 1 hank 

The term "hank" in the table does not mean that the 
yarn is in hank form, as it may be in any of the usual 
forms, such as cops, Ijobbins, warps, skeins, or hanks, 
etc., but it is used to indicate a definite length of yarn, 
i. e. 840 yards. 

The table of weights for Cotton Yarn is as follows : 
437i grains ^ 1 oz. avoirdupois, 
7000 " = 16 oz. = 1 lb. 

The above table is the standard for weight used for 
for all materials except raw silk. The table of lengths 
given is the standard for cotton in the United States 
and England. 

Cotton Yarns are indicated by numbers or counts, 
and the number of hanks of 840 yards that weigh 1 lb. 
is the count of the varn. If it should be found that 



2520 yards of a coiton yarn weiglied 1 lb. there 
would be 

2520 ^ 840 = 3 hanks, 

and the counts of the yarn would be 3s. 

Therefore the following rule may be used : 

To find the counts of a Cotton Yarn when the 
length and weight are known : 

Rule. Divide the length by the standard, 840, re- 
sult equals hanks : divide hanks by weight in lbs., equals 
counts. 

Examples. 

If 16800 yards of Cotton Yarn weigh 1 lb., find the 
counts. 

16800 ^ 840 = 20 hanks -^ 1 lb. = 20s counts. 

If 33600 yards of Cotton Yarn weigh 5 lbs., what 
are the counts ? 
33600 -^ 840 = 40 hanks -i- 5 lbs. = 8s counts. 
If 80 bundles of yarn each contain 2100 yds. and 
weigh 20 lbs.- in all, find the cotton counts. 
2100 X 80 = 168000 yds. 
168000 -^ 840 = 200 hanks. 
200 -^ 20 = 10s counts. 
If a warper beam contains 420 ends, 1'8000 yards 
long, and weighs 600 lbs. net, what are the cotton 
counts ? 

18000 X 420 = 7560000 yds. 
7560000 -f- 840 = 9000 hanks. 
9000 ^ 600 z= 15s counts. 

If a warper beam contains 8 wraps of .500 ends and 
weighs 700 lbs. net, what are the cotton counts? (A 
standard wrap is equal to 3000 yards). 

500 X 3000 X 8 



840 X 700 



20.40s counts. 



If a set of 8 section beams of 7 wraps and 450 ends 
each, weigh 750 lbs. in all, find the cotton counts. 

450 X 8 X 3000 X 7 ^^^ 

:= 120s counts. 

840 X 750 

If a loom beam of 2400 ends and 20 cuts of 50 yds. 
each, with 10% size, weighs 45 lbs., what are the cot- 
ton counts of the yarn ? 

45_10% = 40.50 lbs. net. 
20 X 50 X 2400 



840 X 40.50 



70.54s counts. 



The preceding examples show that yarn in any form 
may be reduced to yards, and by dividing by 840 be 
changed to hanks ; by dividing the hanks by the weight 
the counts of the yarn is obtained. 

REELING YARN. 

The counts of yarn on bobbins, cops, or spools, etc., 
may be found by reeling 120 yards on a wrap reel and 
dividing its weight in grains into 1000. 

Example. If 120 yards or 1 skein of Cotton Yarn 
weighs 20 grains, what are the counts ? 
1000 ~ 20 = 50s counts. 

A full hank is too much to reel, so 1 skein, or one- 
seventh of a hank is taken, and its weight divided into 
1000, which is one-seventh of a lb. of 7000 grains. 

If 2 skeins, or 240 yards, are reeled, 2000 must be 
used for a dividend, as it is the same proportion of 
7000 as 240 yards is of a hank. If 4 skeins or 480 
yards are weighed, then 4000 is tlie dividend, etc. 

It is the practice in mills to test the yarn directly 
from the mules or spinning frames at least once a day, 
to see if the yarn is of the correct counts. These tests 



8 

are made by taking 4 cops or bobbins of each count of 
yarn being spun and winding 120 yards, or 1 skein, 
from each on a wrap reel, weighing each skein sepa- 
rately, and dividing its weight in grains into 1000. 
The result is the counts of the yarn in each skein, and 
shows the variation in the counts of the yarn. These 
separate counts may then be added together and the 
result divided by 4 to find the average counts, or the 
4 skeins may be weighed together and their weight in 
grains divided into 4000, the result l)eing the counts, 
as follows : 

20 = 50s counts. 
1000 



1 skein 20 grains, 1000 



1 


u 


20.5 


1 


u 


19.5 


1 


u 


19 



20.5= 48.78 
1000 ~ 19,5 =r 51.28 
1000 -^ 19 = 52.63 



202.69 
202.69 -f- 4= 50.67 average counts. 

Examples. 

If 4 skeins are reeled and together they weigh 
200 grains, what are the counts ? 

4000 -f- 200 = 20s counts. 

If 1 skein of cotton yarn weighs 20 grains, find the 
counts. 

1000 -^ 20 = 50s counts. 

If 2 skeins of cotton yarn weigh 35 grains, what are 
the counts ? 

2000 ^ 35 = 57.14 counts. 

If 4 skeins are wrapped and together weigh 60 
grains, find the counts. 

4000 ^ 60 = 66.66 counts. 

To find the Cotton counts from irregular or short 
lengths of yarn: 



Rule. Multlyly the numher of yards weighed by 8i 
and diride the result by their weight in grains. 

The constant 8^ is the weight in grains of 1 yard of 
number 1 cotton yarn, and is found l)y dividing the 
standard for weight, 7000 grains, by the standard for 
length, 840 yards. 

7000 ^ 840 = 8i grains, weight of 1 yard of num- 
ber 1 cotton yarn. 

Examples. 

If 96 yards of cotton yarn weigh 20 grains, what are 
the counts ? 

As 1 yard of number 1 yarn weighs 8i grains, 96 
yards will weigh 

96 X 81 = 800 grains, 
and as the 96 yards in the example weigh 20 grains, it 
will be 

800 -f- 20 = 40 times lighter than a number 1 or 
40s yarn. 

If 66 yards of cotton weigh 12 grains, what are the 
counts ? 

66 X 8^ = 550 grains. 
550 -^ 12 = 45.83s counts. 

If 135 yards of cotton yarn weigh 20 grains, what 
are the counts ? 

^ -^ = 56.25s counts. 



20 

If 240 yards of cotton yarn weigh 50 grains, what 
are the counts ? 

240 yards = 2 skeins. 

2000 -^ 50 = 40s counts, or 

240 X 81 
^ ^ 40s counts. 



50 



10 

Another rule that will be found useful when only- 
short lengths of thread are available is as follows: 

Rule. TJie number of inches of cotton yarn that weigh 
1 grain, multiplied by .2314, equals the counts. 

Example. If 63 threads, 3 inches long, weigh 1 
grain, what are the cotton counts ? 

63 X 3 = 189 inches of yarn that weigh 1 grain. 
189 X .2314 =r 43.73s counts. 

The constant .2314 is the weight ii> grains of 1 inch 
of number 1 cotton yarn ; hence, any number of inches 
that weigh one grain will be .2314 times finer than 
number 1 yarn. 

840 X 36 = 30240 inches in one hank of cotton 
yarn. 

7000 grains -^ 30240 = .2314 grains, weight of 1 
inch of number 1 yarn. 

This same problem may be worked out by reducing 
the inches to yards and using the constant 8i as follows : 

63 X 3 = 189 inches. ' 

189 -f- 36 = 5.25 yards. 

5.25 X 81 

^ m 43.73s counts. 

1 

To find the length in yards when the counts and 
weight are known : 

Rule. Multiply the standard length, 840 yards, by 
the counts, result equals yards in 1 /6., multiply by weight 
in lbs., equals length. 

Examples. 

Find the length of yarn in 10 lbs. of 20s cotton yarn. 

840 X 20 = 16800 yards in 1 lb. of 20s yarn. 
16800 X 10 = 168000 yards in 10 lbs. 



11 

What is the length contained in 25 lbs. of 40s cot- 
ton yarn ? 

840 X 40 X 25 = 840000 yards. 

If a bobbin or spool filled with yarn weighs 1 lb. 4 
oz., tlie empty bobbin or spool 6 oz,, and the counts of 
the yarn 36s, what length of yarn is contained on the 
bobbin or spool ? 

1 lb. 4 oz. rrr 20 OZ. 

20 — 6 = 14 oz., weight of yarn. 
14 X 437.5 = 6125 grains weight of yarn. 
6125 H- 7000 == .875 lbs. 
840 X 36 X .875 = 26460 yds. on bobbin or spool. 

If a warper beam contains 420 ends of 15s yarn, 
and weighs 600 lbs. net, what is the length of yarn on 
the beam ? 

840 X 15 X 600 = 7560000 yards of continuous 
thread. 

7560000 ^ 420 ends ^ 18000 yards. 

If 32 cops of 55s cotton yarn w^eigh 1 lb. and each 
cop contains an equal amount of yarn, what is the 
length contained on each cop ? 

840 X 55 = 46200 yards in 1 lb. of 55s yarn. 
46200 -^ 32 cops = 1443.75 yards on each cop. 

How many yards in a i lb. of 60s cotton yarn ? 
840 X 60 = 50400 yards in 1 lb. of 60s. 
50400 ^ 2 = 25200 yards iu i lb. 

To find short lengths when counts and weight in 
grains are known : 

Rule. MuUijjIy the weight in grains by the counts 
and divide the result bij 8^. 

For convenience, the decimal 8.33 may be used 
instead of the common fraction 8^ ; although 8.33 is 



12 

not the exact figure it is near enough for practical 
purposes. 

Example. How many yards of 40s cotton yarn will 
weigh 36 grains ? 

36 X 40 _ ^ 
= 172.8 yards, or 

^ili-°= 172.86 yards. 
8.33 ^ 

To find weight when length and counts are known: 

Rule. Divide the length of yarn in yards by the 
sta7idard 840, result equals hanks; divide hanks by 
counts, equals weight in lbs. 

Examples. 
Find the weight of 168000 yards of 20s cotton yarn. 

168000 ^ 840 = 200 hanks. 
200 -^ 20s r= 10 lbs. 

What is the weight of 18480 yards of 44s cotton 
yarn? 

18480 ^ 840= 22 hanks. 
22 -^ 44 = .5 or i lb. 

If a warper beam contains 500 ends of 32s cotton 
yarn and is 24000 yards long, what is the weight of 
the yarn ? 

24000 X 500 = 12000000 yards. 
12000000 -^ 840 = 14285.71 hanks. 
14285.71 -f- 32 = 446.42 lbs. 

If you were running 410 ends of 36s yarn on your 
warpers, what should 6 wraps weigh ? 

410 X 3000 X 6 



840 X 36 



= 244.04 lbs. 



13 



Find the weight of 27720 yards of 44s cotton yarn, 
= . <5 or 4 II). 



840 X 44 



To find the weight of short lengths of Cotton 
Yarn: 

Rule. Multiply the length by 8^, result equals the 
weight in grains of that length of namher 1 Cotton Yarn; 
divide by any other counts, equals the weight in grains of 
that counts. 

Examples. 

Find the weight of 144 yards of 50s Cotton Yarn. 

144 X 81 = 1200 grains, weight of 144 yards of 
50s Cotton Yarn, and as 50s is 50 times lighter than 
number 1, 

1200 -^ 50 = 24 grains. 

Find the weight of 75 yards of 60s Cotton Yarn. 

75 X 81 



60 



= 10.41 srrains. 



What is the weight of 96 yards of a 24s Cotton 
Yarn ? 

^^-^^ = 33.33 grains. 



SPUN SILK. 

Spun silk in single yarn is numbered by the same 
method as Cotton Yarn, the same table of lengths and 
weights being used. Thus the rules given for calcu- 
lating single Cotton Yarn will apply to spun silk. Ply 
yarn, however, is indicated in a different manner. 

Cotton ply yarn is indicated by placing the number 
of the threads twisted before the numbers of the single 



14 

yarn ; a two ply 40s would be composed of two single 
ends of 40s and would be indicated thus, 2 /40s, and 
would be equal in weight to a single 20s. In spun 
silk, the method is to place the actual size or counts of 
the ply thread first, and the number of threads twisted 
after it. Thus a two ply thread of spun silk equal to. 
a 20s would be composed of two threads of 40s and 
would be indicated 20s/2 ; a 30s/3 would be 
composed of three threads of 90s. 

Thus the count of a ply thread in spun silk, multi- 
plied by the number of threads twisted, equals the 
counts of the single threads. 

Examples. 

If 302400 yards of spun silk weigh 9 lbs., what are 
the counts ? 

302400 

— z=i 40s counts. 

840x9 

If 115 yards of spun silk weigh 20 grains, find the 
counts. 

3 _ 47 90s counts. 

20 

Find the length of 25 lbs. of 40s spun silk. 
840 X 40 X 25 = 840000 yards. 
What counts of spun silk are equal to a 60s cotton ? 
The same counts, 60s. 

What counts of single yarn would be used to make 
a 20s/^3 spun silk thread ? 

20 X 3 = 60s counts. 

How many yards of ply yarn are contained in 10 
lbs. of 20s/2 spun silk ? 

840 X 20 X 10 = 168000 yards. 



15 

How many yards of single yarn would be required 
to make 10 lbs. of 20s/2 spun silk? 

20s/2 is made by doubling single 40s. 

840 X 40 X 10 = 336000 yards of single yarn 
required. 

How many yards of ply yarn are contained in 20 
lbs. of 30s/3 spun silk ? 

840 X 30 X 20 = 504000 yards. 

How many yards of single yarn would ])e contained 
in 20 lbs. of 30s/3 spun silk? 

30s/3 would be made of 90s single. 
840 X 90 X 20 = 1512000 yards. 

RAW SILK. 

Silk is the product of the silk worm, which, at a 
certain stage of its existence, discharges the silk in a 
semi-fluid condition from two glands near its head. 
These filaments unite as they are discharged and form 
one thread which hardens immediately on exposure to 
the air. 

The worm winds the thread around itself, so that 
when it has finished it is entirely enclosed in what is 
called a cocoon, which contains on a average about 
400 yards of thread. 

In about three weeks, after it has finished the 
cocoon, the worm changes to a moth which forces its 
way out of the cocoon, cutting or breaTcing some of the 
threads which spoils the cocoon. To prevent this, 
every cocoon, not intended for breeding purposes, is 
placed in a steam heater to stifle the chrysalis. 

The silk may then be reeled at any future time. 

Four or more of these cocoons (according to the 
size of thread desired) are placed in a basin of warm 



16 

water to soften. Then the threads are slightly twisted 
around themselves, run on a reel and made into skeins. 

The skeins weigh from one to several ounces and 
are packed into bundles called books, weighing from 
five to ten pounds. The silk is then made up into 
bales weighing from one hundred to one hundred and 
sixty pounds weight and is called thrown silk. 

When preparing thrown silk for weaving purposes, 
it is sometimes necessary to twist several threads 
together. The threads to be used as warp are twisted 
harder than those for filling and will now be known as 
organzine, while those for filling will be known as 
tram. 

As in the case of Cotton Yarn, the warp threads are 
generally coarser than the filling threads and this is 
obtained by twisting more threads together for the 
warp than for the filling. 

A 4 thread organzine means that 4 threads of 
thrown silk, (raw silk) have been twisted sufficiently 
hard for warp, while if 3 threads were twisted with a 
less number of turns per inch for filling it would be 
called tram. 

Raw silk is numbered on an entirely diff'erent basis 
from either spun silk, cotton, worsted, or woolen yarn. 
In the United States and in England, the length of the 
hank used is 1000 yards, and the number of drams 
that such a hank weighs is the counts of the silk. 
Thus, in raw silk, the coarser the silk the higher the 
counts. 

For calculating raw silk in the United States and 
English system the following table of weights is used : 

1 dram = 27.34 grains. 

16 " =1 oz. = 437.5 
256 " = 16 '^ z= 7000 " = 1 lb. 



17 

The table of length is : 

1000 yards equals 1 liaiik. 

Thus, if 1000 yards weigh 1 dram, it would be a 1 
dram silk, and as there are 256 drams in 1 lb., there 
will be 1000 X 256 := 256000 yards in 1 lb. of 
number 1 dram silk. 

To find the dram silk counts when length and 
weight in lbs. are known : 

Rule. Multiply 256 ^y/ 1000, result, equals yards of 
1 dram silk in 1 lb.; multiply by iveight in lbs., equah 
total length; divide by yards iveighed, equals counts. 

Examples. 

If 64000 3^ards of raw silk weigh 2 lbs., find the 
dram silk counts. 

In 2 lbs. there are 512 drams and 512000 yards of 
1 dram silk, and, as the 64000 yards in the example 
weigh 2 lbs. or 512 drams, they are: 
512000 ^ 64000 = 8 times heavier or 8 dram silk. 

If. 128000 yards of raw silk weigh 3 lbs., what are 
the dram silk counts ? 

256 X 1000 X 3 ^ ^ 

—- 6 dram silk. 

128000 

If 512000 yards of raw silk weigh 10 lbs., what are 
the dram silk counts ? 

256 X 1000 X 10 

_ = 5 dram silk. 

512000 

To find length when counts and weight are known: 

Rule. Midtiply 25Q by 1000, result equals yards of 

1 dra?n silk in 1 lb.; midtiply by weight in lbs., equals 

total length ; divide by the counts, equals length of that 
count of silk. 



18 

Examples. 
Find the length in 5 lbs. of 4 dram silk. 

256 X 1000 X 5 = 1280000 yards in 5 lbs. of 1 
dram silk. 

A 4 dram silk is 4 times heavier than a 1 dram silk, 
consequently there will be 4 times less length in the 
same weight of a 1 dram silk. 

1280000 ^ 4 = 320000 yards. 

Find the length in 10 lbs. of 8 dram silk. 

256 X 1000 X 10 

= 320000 yards. 

8 ^ 

Find the length in 5 J lbs. of 6 dram silk. 

256 X 1000 X ^5.5 ^_^^ _ 

= 234666.66 yards. 

6 ^ 

To find the weight when the length and counts 
are known : 

Rule. Multiply 25Q bij 1000, result equals yards of 
1 dram silk in 1 lb.; divide by the counts, equals yards 
in 1 lb. of that counts; divide total yards by the yards jier 
lb., equals weight in lbs. 

Examples. 

Find the weight of 160000 yards of 8 dram silk. 

256000 yards of 1 dram silk weigh 1 lb., and as an 
8 dram silk is 8 times heavier than a number 1, there 
will be 

256000 -^ 8 = 32000 yards of 8 dram silk in 1 
lb., and 32000 is contained in 160000 5 times, which 
equals 5 lbs. 

Find the weight of 234667 yards of 6 dram silk. 

256 X 1000 ^ 42666.66 yards of 6 dram silk in 1 lb. 
6 ^ 

234667 -^ 42666.66 = 5.5 lbs. 



19 

The same result may be obtained by another 
method as follows : 

Rule. Multiyly the yards hy the counts, result equals 
the length in 1 dram co2mts, divide by 256 X 1000. 
equals the weight in lbs. 

The last example by this rule is as follows : 
234667 X 6 



256 X 1000 



5.5 lbs. 



To find counts when length and weight in ounces 
are known : 



Rule. Multiply freight in ounces by 16, result equals 
iveight in drams ; divide length by drams, equals yards 
jmr dram; divide 1000 by yards per dram, equals counts. 

Example. If 2400 yards of raw silk weigh 3 
ounces, what are the dram counts ? 

16 X 3 = 48 drams weight. 
2400 ^ 48 = 50 yards per dram. 
1000 ^ 50 = 20 dram silk. 

To find length when counts and weight in ounces 
are known : 

Rule. Multiply the iveight in ounces by 16, result 
equals weight in drams; divide 1000 by counts, equals 
yards jier dram ; multiply iveight in drams by yards per 
dram, equals length. 

Example. Find the length of raw silk in 7 ounces 
of 10 dram silk. 

16 X 7 = 112 drams weight. 
1000 -f- 10 =100 yards per dram. 
112 X 100= 11200 yards. 



20 

To find weight in ounces when length and counts 
are known : 

Rule. Divide loot) bij the counts, result eqidcils yards 
per dram ; divide total yards by yards per dram, ecjuals 
weight in drams ; divide iveight in drams by IQ, e(pials 
iveight in ounces. 

Example. What is the weidit in ounces of 2000 
yards of 12 dram silk? 



1000 

2000 

24 



12 ^ 83.33 yards per dram. 
83.33 = 24 drams weight. 
16 := 1.5 ounces weidit. 



■■o' 



To find the dram silk counts when length and 
weight in grains are known : 

Rule. Divide weight in grains by 27.34, result 
equals weight in drams ; divide length by iveight in 
drams, equals yards per dram ; divide 1000 by yards per 
dram, equals counts. 

Example. If 1280 yards of raw silk weigh 70 
arrains, what are the dram silk counts ? 



70 -^ 


27.34 = 2.56 drams weight. 


1280 ^ 


2.5.6 = 500 yards per dram 


1000 ^ 


500 = 2 dram silk. 



To find length when counts and weight in grains 
are known : 

Rule. Divide weight in grains by 27.34, residt 
equals weight in drams; divide 1000 bij counts, equals 
yards per dram; multiply iveight in drams by yards per 
dram, equals length. 

Example. Find the length of raw silk in 100 
grains weight of 4 dram silk. 

100 -^ 27.34 = 3.65 drams weight. 
1000 ^ 4 =: 250 yards per dram. 
3.65 X 250 = 912.50 yards. 



21 

To find weight in grains when length and counts 
are known: 

Rule. D'tride lOfKJ hy the coaTtn. -^ti'^c t^ju/j'S 
yardi per dram : diride UAal yardt hy yard* per dram. 
eqvaU i«aVA/ in dramjs ; multiply 27-M by wfight in 
dram*. (^piaU tceighi in grains. 

Example. Find weight in grains of 1500 yards of 
12 dram silk- 

1000 -^12 — >t?j.?jZ yard* per dram. 
1500 ^ 83.33 = 1^ drams weiglit. 
27-34 X 1« = 492.12 grains weight 

In TT2inQe and other coantrie* of Europe, the 
standard length of the hank is 476 metres, or 520 
yards- and the siandard weight is 1 denier. The 
nam1>er of deniers that 1 hank of 520 yards weighs is 
the counts of the silk. 

The table of weights is as follows : 

533i deniers == 1 oz. aroirdupois- 
8533' - r= 16 ^ =1 lb. 

Raw sflk varies in weight to such an extent that in 
numbering it in the denier system it is the practice to 
allow a variation of two numbers, and instead of 
specifying the counts Ijy a single number, two are 
given- In the case of a 15 denier silk, the counts 
would be marked 14/16. meaning 14 to 16. allowing 
a variation of one number each way and the silk would 
be known as 14/16 denier silk: other counts would 
be known as 18/20. 20/22 etc. 

If a calculation were made on a 14/16 denier silk 
the number between the two specified, or 15. would be 
used: on an 18/20. 19 would ]>e used etc. 



22 

To find the denier counts of raw silk, when 
length and weight in lbs. are known : 

Rule. Multiplij 8533 hij 520, result equals yards of 
1 denier silk in 1 lb. ; multiply by weight in lbs., equals 
total length; divide by yards weighed, equals counts. 

Examples. 

If 400000 yards of raw silk weigh 2 lbs., what are 
the denier counts ? 

8533 X 520 X 2 = 8874320 yards in 2 lbs. of 1 
denier silk, and, as the 400000 yards in the example 
weigh 2 lbs:, they are 

8874320^400000 = 22.18 times heavier or 22.18 
denier counts. 

If 320000 yards of raw silk weigh li lbs., what are 
the denier counts ? 

8533 X 520 X 1.5 

— = 20.78 denier counts. 



320000 

To find length when weight in lbs. and counts 
are known : 

Rule. Multiply 8533 by 520, result eq^ials yards oj 
1 denier silk in 1 lb.; multij^ly by weight in lbs., equals 
total length in that weight of I denier silk; divide by the 
counts, equals length. 

Examples. 

How many yards are contained in 4 lbs. of 13/15 
denier silk ? 

8533 X 520 X 4= 17748640 yards of 1 denier 
silk in 4 lbs. weight, and, as a 14 denier silk is 14 
times heavier than a 1 denier, there will be 

17748640 ^ 14 = 1267760 yards. 



23 

How many yards are contained in 7 lbs. of 20 / 22 
denier silk ? 

8533 X 520 X 7 

1^ = 1479053 yards. 

21 ^ 

Find the length in 10 lbs. of 14/16 denier silk. 

8533 X 520 X 10 

= 2070674.66 yards. 

15 ^ 

To find weight when length and counts are 
known : 

Rule. Multiply 8533 by 520, resiik equals yards of 
1 denier silk in 1 lb.; divide by the counts, equals yards 
in 1 lb. of that counts; divide total yards by yards per 
lb., equals weight in lbs. 

Example. Find the weight of 450000 yards of 
15/17 denier silk. 

8533X 520 

~ =277322 yards of 16 denier silk in 1 lb. 

16 

450000 -^ 277322 = 1.62 lbs. weight of the 450000 
yards. 

The same result may be obtained by another 
method as follows : 

Rule. Multiply the yards by the counts, result equals 
the length in 1 denier counts, divide by 8533 X 520, 
equals the weight in lbs. 

The last example by this rule is as follows. 

450000 X 16 

= 1.62 lbs. 

8533 X 520 

Practically all the raw silk used in the United States 
is imported and is numbered in the denier system. 



24 

For^'convenience in calculating according to English 
and United States methods, the denier counts may l)e 
changed to dram silk counts as follows : 

To find the equivalent counts of denier silk in 
dram silk counts: 

Rule. Multiply 8533 by 520, result equals yards of 
1 denier silk in 1 lb. ; divide by the denier counts, equals 
yards in 1 lb. of that counts; divide by 25^000, .the yards 
of 1 dram silk in 1 lb.; and ansiver is dram silk coimts. 

Example. Find the equivalent of a 11/13 denier 

silk in dram silk counts. 

8533X520 

— z= 369763 yards of 12 denier silk in 1 lb. 

12 

369763 -^ 256000 = 1.44 dram silk. 
Or in one operation as follows 

8533 X 520 

=r: 1.44 dram silk. 

12 X 256000 

Thus a 11/13 denier and a 1.44 dram silk are 
equal. 

To find the equivalent counts of dram silk in 
cotton counts : 

Rule. Multiply 256 by 1000, result equals yards of 
1 dram silk in 1 lb.; divide by the dram silk counts, 
equals the yards iii 1 lb. of that counts ; divide by 840, 
equals the counts in the cotton system. 

Example. Find the equivalent counts of a 4 dram 

silk in cotton counts. 

256 X 1000 

= 64000 yards of 4 dram silk in 1 lb. 

4 ^ 

64000 ^ 840 = 76.19 cotton counts, or 

256 X 1000 

= 76.19 cotton counts. 



4 X 840 



25 

Thus a 4 dram silk and a 70 cotton yarn are equal. 

To find the equivalent counts of a denier silk in 
cotton counts : 

Rule. Multiply 8533 hy 520, result e(pials yards of 
1 denier silk in 1 lb.; divide by the dewier silk counts, 
equals the yards in 1 lb. of that counts; divide by 840, 
equals the counts in the cotton system. 

Example. Find the equivalent of a 21/23 denier 
silk in cotton counts. 

^ = 201689 yards of 21/23 denier silk 

22 in 1 lb. 

201689 4- 840 = 240.1 cotton counts, or 
8533 X 520 



22 X 840 



240.1 cotton counts. 



Thus a 21/23 denier silk and ^ 240 cotton yarn 
are equal. 

WORSTED YARN. 

The table of lengths for worsted yarn is as follows : 

1 yard = 1 thread or cir. of worsted reel. 
80 '' = 80 " =1 lea or knot. 
560 '^ = 560 " = 7 " " " == 1 hank. 

The table of weights is the same as that for Cotton 
Yarn. 

The number of hanks of 560 yards contained in 1 
lb. indicates the counts. 

If 560 yards instead of 840 are used, the rules given 
for Cotton Yarn will apply equally well for worsted, 
except those given for dealing with short lengths. 



26 

Examples. 

If 252000 yards of worsted yarn weigh 15 lbs., 
what are the counts ? 

252000 -^ 560 = 450 hanks. 
450 -f- 15 = 30s counts. 

How many yards are contained in 25 lbs. of 20s 
worsted yarn ? 

560 X 20 X 25 == 280000 yards. 
What is the weight of 268800 yards of 40s worsted 



yarn ? 








268800 -f- 


560 = 480 hanks. 




480 -^ 


40s = 12 lbs. 



If a beam that contains 1568 ends, 300 yards long, 
weighs 30 lbs. net, what are the counts ? 
1568 X 300 = 470400 yards. 
470400 ^ 560 = 840 hanks. 
840 -^ 30 = 28s counts. 

To find the worsted counts from irregular or 
short lengths : 

Rule. Multiply the number of yards weighed by 12.5 
and divide the result by their tveight in grains. 

The constant 12.5 is found by dividing 7000 grains 
by 560 yards, length of a worsted hank. 

7000 ^ 560 = 12.5 grains, weight of 1 yard of 
number 1 worsted yarn. 

Examples. 

If 56 yards of worsted yarn weigh 35 grains, what 
are the counts ? 

56 X 12.5 

= 20s counts. 

35 



27 

As 1 yard of number 1 yarn weighs 12.5 grains, 56 
yards will weigh 

56 X 12.5 = 700 grains; and, as the 56 yards in 
the example weigh 35 grains, it will be 

700 -i- 35 = 20 times finer than a number 1 or 
20s counts. 

If 63 yards of worsted yarn weigh 40 grains, what 
are the counts ? 

63 X 12.5 

z= 19.68s counts. 

40 

If 110 yards of worsted yarn weigh 50 grains, what 
are the counts ? 

110 X 12.5 

=^ 27.5s counts. 

50 



WOOLEN YARN. 

There are two standards for length used in the 
calculations for woolen yarn ; viz. the run and the cut 
systems. 

In the cut system, 300 yards is the standard for 
lengtli, and the number of cuts of 300 yards contained 
in I lb. is the counts of the yarn. 

In the run system, 1600 yards is the standard for 
length, and the number of runs of 1600 yards that 
weigh 1 lb. is the counts of the yarn. 

The full table of woolen lengths is as follows : 

1 yard =^ 1 thread or cir. of the woolen reel. 
80 " = 1 knot. 
300 " = 3| ^' = 1 cut. 
1600 '^ =20 ^' = 51 '^ = 1 run. 



28 

To find counts, length, or weight, apply rules 
given for cotton and worsted yarns (except those 
for short lengths) and use the standard length in 
the system desired. 

WOOLEN CUT YARN. 

Examples. 

What are the cut counts of a woolen yarn when 
1200 yards weigh 2 lbs.? 

1200 ^ 300 r=r 4 cuts. 

4 -f- 2 = 2s cut counts. 

If 45 bundles of woolen yarn each contain 600 
yards and weigh 30 lbs. in all, what are the cut counts ? 
600 X 45 = 27000 yards. 
27000 ^ 300 = 90 cuts. 

90 -^ 30 = 3s cut counts. 

Find the length in 100 lbs. of 3 cut yarn. 

300 X 3 X 100 = 90000 yards. 

If 40 spools of 5 cut yarn weigh 6 lbs. net and each 
spool contains an equal amount of yarn, find t?he yards 
on each spool. 

300 X 5 X 6 = 9000 total yards. 

9000 -^ 40 = 225 yards on each spool. 

Find the weight of 60000 yards of 4 cut yarn. 
60000 -^ 300 = .200 cuts. 
200 ^ 4 = 50 lbs. 

If a loom beam contains 1400 ends of 6 cut yarn 
and is 500 yards long, find the weight of the yarn. 
1400 X 500 =3 700000 total yards. 
700000 ^ 300 = 2333.33 cuts. 

2333.33 ^ 6 = 388.88 lbs. weight of yarn. 



29 

To find the cut counts from short lengths : 

Rule. Multlpbj the jjards weighed by 23^ and divide 
result by weight in grains. 

Example. If 45 yards of woolen yarn weigh 240 
grains, wliat are the cut counts ? 

45 X 23.33 

= 4.3 i cut counts. 

240 

The constant 23^ is the weight in grains of 1 yard 
of a 1 cut yarn and holds the same relation to the 300 
yard cut that 8i does to the 840 yard hank, and 12.5 
does to the worsted hank of 560 yards. 

WOOLEN RUN YARN. 

Examples. 
If 100000 yards of woolen yarn weigh 7^ lbs., what 
are the run counts ? 

100000 -^ 1600 =rr 62.5 runs. 

62.5 ^ 7.5 ^ 8.33 run yarn. 

Find the length of 75 lbs. of 2 run woolen yarn. 

1600 X 2 X 75 = 240000 yards. 
What is the weight of 38400 yards of 3 run woolen 



yarn ? 








38400 ^ 


1600 = 24 runs. 




24 ^ 


3 = 8 lbs. 



To find «ounts from short lengths: 

Rule. Multiply the yards weighed by 4.375 and 
divide result by weight in grains. 

If 60 yards of woolen yarn weigh 20 grains, find 
the run counts. 

60X4.875 ,„,„ 

=r Id. 12 run counts. 

20 



30 

In the run system, the counts equal the number ot 
hundred yards in an ounce of the yarn. 

1600 yards in 1 run divided by 16 oz. per Ib.=:100 
yards. 

Thus, to find the counts when the weight is 
known in ounces : 

Rule. Divide the length by 100 and divide the result 
by the number of ounces the yarn weighs. 

Examples. 

If 13200 yards of woolen yarn weigh 33 ounces, 
what are the run counts ? 

13200 ^ 100 =z 132. 

132 -f- 33 ^ 4 run yarn. 

If 430 yards of woolen yarn weigh 1^ ounces, what 
are the run counts ? 

430 -f- 100 = 4.30. 
4.30 -^ 1.5 = 2.86 run yarn. 

To find the weight in ounces when the length and 
run counts are known : 

Rule. Divide the length by 100 and divide the result 
by the counts of the yarn. 

Examples. 

Find the weight in ounces of 6600 yards of a 4 
run yarn. 

6600 -f- 100 =: 66. • 

66 -i- 4 = 16.5 ounces weight. 

Find the weight in ounces of 320 yards of 3 run 
yarn. 

320 ^ 100 = 3.2. 
3.2 ^ 3 = 1.06 ounces weight. 



31 

To find the length when counts and weight in 
ounces are known : ' 

Rule. Multiply the iceight in ounces by 100 and 
multiply this result by the counts of the yarn. 

Example. Find the length of 27 ounces of 6 run 
woolen yarn. 

27 X 100 X 6 = 16200 yards. 

LINEN YARN. 

The length of the linen hank is 300 yards, and the 
standard of weight is 1 ll>. The number of hanks that 
weigh 1 lb. is the count of the yarn. The rules given 
for calculating woolen yarn in the cut system apply to 
linen yarn also. 

Cotton, in process of manufacture, is worked tlirough 
the pickers, cards, and drawing frames, witliout Iiaving 
twist put into it, but from the slubber to the spinning 
frame or mule the amount ot twist put in increases with 
every process. Twist is put in roving and yarn to give 
it strength, and is calculated by the number of turns 
put in one inch. 

The turns per inch vary according to the quality ot 
the cotton used and the class of yarns being made, and 
is determined by multiplying the square root of the 
counts by a constant. 

The constant for warp yarn in the U. S. is 4.75. 
" " " filling " " 3.25. 

" " '' hosiery" " 2.50. 

To find the turns of twist per inch : 

Rule. Multiply the square root of the counts by one 
of the constants. 



32 

Examples. 

Find the turns per inch for a 25s warp yarn, using 
4.75 for a constant. 

V25 = 5, the square root of 25 is 5. 
5 X 4.75 = 23.75 turns per inch. 

Find the turns per inch for a 20s hosiery yarn. 

\/20~= 4.47. 
4.47 X 2.50 r= 11.17 turns per inch req. 

Warp yarns are tested for strength by breaking one 
skein on a testing machine that registers the breaking 
weight in lbs. 

To find the standard breaking weight of warp 
yarn : 

Rule. Divide 1760 by the counts of the yarn; result 
is the St an do rd hrcakinor weiorht in lbs. 

b o 

Example. What is the standard breaking weight of 
a 40s cotton warp yarn ? 

1760 -^ 40s = 44 lbs. 

Although 1760 is accepted as the standard constant 
for breaking weight, it will be found in cloth mills, 
where medium and finer counts are woven, that a 
higher constant must be used to get satisfactory results 
in the weave room, while in mills on coarser counts, a 
lower constant may be used and satisfactory results 
follow. 

The following table of constants has been compiled 
partly from Draper's table of brCakinsi; weights and 
from the writer's personal experience in the weaving 
of fine counts. 



33 



Counts. 


Constants, 


20 to 29 


1760 


30 - 39 


1780 


40 '• 49 


1825 


50 '^ 59 


1880 


60 -' 69 


1930 


70 '' 79 


1960 


80 ^' 89 


1980 


90 " 99 


2040 


100 '• 109 


2150 


110 '' 119 


2200 



Constant divided by warp counts equals the lbs. that 
one skein of cotton yarn should register on the breaking 
machine. 

EQUIVALENT COUNTS. 

When dealing with the numbering of diiferent yarns, 
it frequently becomes necessary to know what the 
counts of a certain yarn would be if numbered accord- 
ing to a different standard. 

This is known as converting tlie counts of yarn in 
one system into equivalent counts of another system. 

To find the equivalent counts of a yarn in any 
other system (except raw silk) : 

Rule. Mtdtijfhj the given, counts hij its own standard'' 
length, result equals yards in 1 lb., divide by the standard/ 
yards in the system desired, equals the counts. 

Examples. 

What are the equivalent counts of a 15s cotton in 
woolen run counts ? 

840 X 15 = 12600 yards in 1 lb. of 15s cotton 
yarn. 



34 

12600 ^ 1600 yards in 1 run woolen = 7.87 run 
counts. 

Find the equivalent counts of a 50s worsted in 
cotton counts. 

560 X 50 



840 



33.33 cotton counts. 



Find the equivalent of a 40s/ 2 spun silk in worsted 
counts. 



840 X 40 
560 



60s worsted. 



Find the equivalent of an 8 cut woolen in run 

counts. 

300 X 8 ^ ^ 

=1.5 run counts. 

1600 

PLY YARNS. 

Ply yarns are made by twisting two or more threads 
of the same counts of single yarn, and are then 
numbered according to the counts of the single yarn 
with the number of threads twisted put betore it 
(except in spun silk). Thus, if two threads of 40s 
were twisted, it would be numbered 2/^40s, and would 
equal a 20s single ; if three threads of 35s were twisted 
it would be numbered 3/ 35s, and would equal a 
11.66s single etc. Finding the counts of the yarn after 
they have been twisted is known as finding the re- 
sultant counts. 

RESULTANT COUNTS. 

It sometimes occurs in fancy yarns that threads of 
unequal counts are twisted together, and in this 
instance, the resultant count must be found in another 
way than when the counts of the threads are equal. 



35 

To find the resultant counts of a ply thread when 
two or more threads of different counts are twisted 
together : 

Rule. Dir'ule the highest counts by \t self and by each 
of the other counts; add the results, and- dirlde into the 
highest counts. 

Examples. 

Suppose one thread of 60s and one of 20s worsted 
yarn are twisted together ; what is the resultant counts 
of the twisted thread ? 

Assuming there is no variation in the contraction ot 
either thread during the twisting, the same length of 
each counts will be required. 

Suppose that 60 hanks or 1 11). of the 60s were used, 
then 60 hanks of the 20s would also be used, and when 
these have been twisted together there will be 60 hanks 
in length of the combined threads. But as 60 hanks 
of 60s weigh I lb., 60 hanks of 20s will weigh 3 lbs.; 
consequently the 60 hanks of twisted thread will weigh 
4 lbs. and 

60 -i- 4 = 15 hanks per lb. or 15s resultant counts. 
This may be stated as follows : 

60 hanks of 60s = 1 lb. 
60 " '^ 20s = 3 lbs. 

4 lbs. total. 

60 lianks of twisted thread now weigh 4 lbs., and 

60 -f- 4 = 15s resultant counts. 

This same method can be used when more than two 
threads are twisted. 



36 

Example. Find the resultant counts when a 40t 
20s, and 10s cotton threads are twisted together. 
40 -^ 40 = 1 lb. 
40 -^ 20 rr= 2 - 
40 -^ 10 = 4 " 



7 lbs. total. 
40 hanks of twisted thread now weigh 7 lbs. and 

40 ^ 7 = 5.71 resultant counts. 
When ply yarns are composed of difterent materials 
it is first necessary to reduce all the yarns to one 
standard, (see equivalent counts) and then proceed as- 
by last rule. 

Example. If a ply thread is composed of one 
thread of 24s worsted, one thread of 16s worsted, and 
one thread of 32s spun silk, what are the resultant 
counts in worsted ? 

840 X 32 ^^ 

m 4bs worsted. 

560 

Tlie equivalent of a 32s spun silk in worsted counts 
is 48s ; thus we have one thread of 48, one of 24, and 
one of 16s worsted; then by rule, 

48 -^ 48 = 1 lb. 
48 ^ 24 = 2 " • 
48 ^ 16 =^ 3 " 
48 hanks weigh 6 lbs. 
48 hanks -^ 6 lbs. = 8 hanks per lb., or 8s re- 
sultant counts. 

To find the cotton counts of yarn required to 
twist with a known count of single yarn, to produce 
a given count of ply yarn : 

Rule. Divide 840 by the difference in weight of the 
known single and ply yarns. 



37 

Example. What counts of cotton yarn is required 
to twist with a 40s cotton yarn to make a ply yarn 
equal to a 15s ? 

840 -^ 15s = 56 lbs. weight of 840 hanks of 15s. 

840 ^ 40s =21 '^ " '• ^' '' '• 4:0s. 

35 lbs. diflference. 
840 hanks 4- 35 lbs. = 24s counts required. 

It has previously been explained that the same 
length of each counts of single yarn will be required 
to make a ply yarn but that the weights will be 
different. 

In this case, 840 hanks is taken so as to have the 
weights show in lbs., any other length could be used 
and tlie answer obtained the same way ; as for instance, 
if 1000 yards were used: 

1000 ^ 15s = 66.66 represents total weight. 

1000 — 40s ~ 25 " weight of 40s. 



41.66 represents difference in wgt. 



1000 ^ 41.66 = 24s counts required. 

To find the weight of each counts of yarn to 
obtain a given weight of ply yarn, when the threads 
twisted are of unequal counts : 

Rule. First find the counts of the yhj yarn, then 
multiply by the iveight of ply yarn required, divide the 
result by the counts of either of the single yam, and the 
result equals the iveight of that yarn ; deduct this iveight 
from the total iveight and the remainder is the iveight of 
tlie other counts of single yarn. 

Example. What weight of 40s cotton yarn must 
be twisted with a 32s cotton yarn to produce 100 lbs. 
of ply yarn ? 



38 

40 -f- 40 = 1 lb. 

40 -^ 32 = 1.25 '' 

40 hanks weigh 2.25 lbs. 
40 -i- 2.25 = 17.77 counts of the ply yarn. 
17.77 X 100 lbs. = 1777 hanks in 100 lbs. 
1777 ^ 40s = 44.44 lbs. of 40s required. 
100 lbs. — 44.44 = 55.56 lbs. of 32s. 

It will be plain that the length being the same, the 
weight of the co irser yarn will be greater than the 
finer counts; 17.77 counts of the ply yarn means that 
17.77 hanks weigh 1 lb., and when multiplied by 100 
equals the hanks in 100 lbs. ; as the lengths of the 40s 
and 32s that compose the ply thread are equal, the 
hanks divided by either of the single counts will give 
its weight in that length etc. 

The rule applies when more than two threads are 
twisted. 

Example. Find weight of each of 120s, 80s, and 
40s required in making up 100 lbs of ply yarn. 
120 -f- 120 = 1 lb. 
120 -f- 80 = 1.5 ^' 
120 -^ 40 =z 3 ^' 



120 hanks weigh 5.5 lbs. 
120 ^ 5.5 = 21.82 counts of the ply yarn. 
21.82 X 100 = 2182 hanks of ply yarn in 100 \ht 
weight. 

2182 -^ 120s = 18.18 lbs. of 120s. 
2182 -^ 80s = 27.27 " • -' 80s. 
2182 ^ 40s = 54.55 '^ " 40s. 



100.00 " 



Or multiply the total weight of yarn by the relative 
weight of each count, and divide the result by the sum 
of the relative weights of all tlie yarns. 



39 

The last example by this rule will be 
100 lbs. X 1 



5.5 
100 X 1.5 
5^5 ~ 
100 X 3 
5^5 



= 18.18 lbs. of 120s. 

= 27.27 lbs. of 80s. 

^ 54.55 lbs. of 40s. 



100.00 ^' 



COST OF PLY YARNS. 

To find the cost of a ply yarn when the threads are 
of equal counts, material, and value, is simply to 
multiply the lbs. of ply yarn by the cost of the single 
yarn plus the cost for twisting. 

In the manufacture of fancy yarns, the counts, ma- 
terials, and cost per lb. of the yarns twisted are often 
different. In this case, the price of the finished yarn 
must be found in some other way. 

To find the cost per lb. of a ply yarn composed of 
threads of different counts and values: 

Rule. Divide the highest counts by itself, and then by 
each of the other counts; the result in each case will repre- 
sent the relative weight of each thread in lbs. ; multiply 
the relative weight (f each thread by its own cost ; then 
divide the total cost by the sum of the relative weights; the 
answer equals the cost per lb. of the ply thread. 

Examples. 

If a 10s cotton thread at 15c. and a 40s cotton 
thread at 34c. per lb. are twisted" together, find the 
cost per lb. of the ply thread. 



40 -^ 


- 40 = 1 lb. of 40s X 34c. 


40 -^ 


:- 10 = 4 " '' 10s X 15c, 



40 

34c. 
_ 60c. 

5 lbs. of ply yarn cost 94c. 
94c. -^ 5 lbs. = 18.8c. per lb. 

If the threads are of diflPerent materials and the 
counts calculated on a different standard, reduce them 
to the same standard (see equivalent counts) and then 
proceed as in the last example. 

Find the cost per lb. of a twisted thread composed 
of 1 thread of 40s worsted at 75c. per 11). and 1 

thread of 60s/2 spun silk at $2.30 per lb. 

* 
First reduce the worsted to its equivalent counts in 
spun silk ; 

560 X 40 

=^ 26.66 equivalent count of the worsted 

^^^ in spun silk counts. 

We shall now have yarns as follows : 

A 26.66s at 75c. and a 60s at $2.30 per lb. ; then 
by the last rule : 

60 -^ 60 = 1.00 lb. of 60s at $2.30 = $2.30 

60 ^ 26.66 = 2.25 " " 26.66s at 75c. = 1.69 

3.25 lbs. of ply yarn cost $3.99 

$3.99 -f- 3.25 lbs. = $1.22 cost per lb. of tlie ply 

yarn. 

If three or more threads are twisted together, first 

find the cost of any two when twisted, as by last rule ; 

then use this ply thread and its cost as though it were 

a single thread, and find the cost of this thread and 

the third when twisted together. 

Find the cost of a ply thread composed of 1 thread 
of 50s worsted at 80c. per lb., 1 thread of 28s cotton 
at 30c., and 1 thread of 20 run woolen yarn at 56c. 
per lb. 



41 

First reduce the worsted to cotton counts and lind 
the cost of the worsted and cotton threads twisted. 

= 33.33 equivalent of 50s worsted in 

^40 cotton counts, then 

33.33 -^ 33.33 = 1.00 lb. of 33.33s at 80c. = 80c. 
33.33^28 =1.19 '• ^' 28s at 30c. = 35c. 

2.19 lbs. of ply yarn cost $1.15 

$1.15 ^ 2.19 lbs. = 52c. cost per lb. of the 
worsted and cotton threads twisted. 

Next find the resultant counts of the two threads, 

33.33 ^ 33.33 = 1.00 lb. of 33.33s 
33.33 ^28 = L19 •' " 28s 

2.19 lbs. of ply yarn. 

33.33 ^ 2.19 =: 15.21 resultant counts. 

Thus we have a yarn equivalent to a 15.21s cotton, 
costing- 52c. per lb., to be twisted with a 20 run 
woolen at 56c. per lb. 

We must now reduce the woolen yarn to cotton 
counts : 

^^^^ ^ ^^ = 38.09 equivalent counts of the 20 
840 
run woolen in cotton counts at 56c. per lb., then to 
follow the rule : 

38.09 -^ 38.09 = 1.00 lb. of 38.09s at 56c. = $ .56 
38.09 ^ 15.21 = 2.50 '• '^ 15.2 Is at 52c. = OO 

3.50 lbs. of ply yarn cost $1.86 

$1.86 ^ 3.50 lbs. == 53c. per lb. cost of the three 
ply yarn. 



42 



EXAMPLES FOR PRACTICE. 

Find the length in 3 lbs. of 40s cotton yarn. 

100800 yards. 

What is the length in 8 lbs. of 60s worsted yarn ? 

268800 yards. 

If 40 cops of 60s cotton yarn weigh 1 lb., and each 
cop contains an equal length, what is the length on 
each cop? 1260 yards. 

Find the length in 10 lbs. of 6 cut woolen yarn. 

18000 yards. 

Find the length of 15 lbs. of 8 run woolen yarn. 

192000 yards. 

If 224000 yards of worsted yarn weigh 10 lbs., 
what are the counts ? 40s counts. 

What is the length in 6 lbs. of 40s/ 2 spun silk ? 

201600 yards. 

If 56 yards of worsted yarn weigh 30 grains, what 
are the counts ? 23.33 counts. 

Find the equivalent of a 4 run woolen yarn in cut 
counts. 21.33 counts. 

Find the number of yards in 9 lbs. of 5 dram silk. 

460800 yards. 

If a loom beam with 3000 ends and 14 cuts of 50 
yards each, with 8% size, weighs 60 lbs., what are the 
cotton counts? 45.29 counts. 

Find the equivalent of a 36s cotton yarn in worsted 
counts. 54s counts. 

How many lbs. of 60s cotton yarn will be required 
to make the following set of section beams — 2 beams 



43 

of 450 ends, 3 beams of 400 ends, and 1 beam with 
375 ends; each beam to contain 6 wraps? 

883.92 lbs. 

Find the equivalent of a 1 dram silk in spun silk 
counts. 304.76 spun silk. 

If 60 yards of cotton yarn weigh 5 grains, what 
are tlie counts? 100s counts. 

Find the weight of yarn on a section beam that 
contains 450 ends of 32s cotton yarn and is 10 wraps 
of 3000 yards long. 502.23 lbs. 

If a section beam contains 8 wraps of 3000 yards, 
500 ends, and weighs 700 lbs. net, what are the cotton 
counts ? 20.40s counts. 

If 80 yards of worsted yarn weigh 50 grains, find 
the counts. 20s counts. 

Find the length in 8 lbs. of 2/ 30s worsted yarn. 

67200 yards. 

Find the weight in grains of 112 yards of 40s cotton 
yarn. 23.33 grains. 

If 75 lbs. of 2/40s worsted yarn is made into a 
warp of 1400 ends, how long will the warp be? 

600 yards. 

If 128000 yards of raw silk weigh 1 lb., what is the 
dram silk counts ? 2 dram silk. 

Find the yards contained in 7 lbs. of 6 dram silk. 

298666.66 yards. 

If 72 threads, each 2 inches long, weigh 2 grains, 
what are the cotton counts? 16.66 counts. 

If 120 yards are wrapped from each of 3 cops, and 
together weigh 50 grains, what are the cotton counts ? 

60s counts. 



44 

If 132740 yards of raw silk weigh 3i lbs., what are 
the dram silk counts ? 6.75 dram silk. 

How many yards of 50s cotton yarn will weigh 40 
grains ? 240 yards. 

Find the equivalent counts of a 40s worsted in 
cotton counts. 26.66s cotton. 

How many yards are contained in 6 lbs. of 16/18 
denier silk? " 1566056.47 yards. 

What is the equivalent of a 60s cotton yarn in 
spun silk counts ? The same, 60s. 

If 400000 yards of raw silk weigh 2 lbs., what are 
the denier counts ? 

22.18 or 21/23 denier counts. 

If 2400 yards of woolen yarn weigh 2 lbs., what are 
the cut counts ? 4 cut yarn. 

Find the weight of 65 yards of 70s cotton yarn. 

8.88 grains. 

Find the equivalent counts of a 10/12 denier silk 
in dram counts. 1.57 dram counts. 

Find the equivalent counts of a 20/22 denier silk 
in cotton counts. 251.54s cotton counts. 

How many yards of single yarn are contained in 7 
lbs. of 60s/2 spun silk?" * 705600 yards. 

Find the resultant counts of a thread when a 60s, 
40s, and 20s cotton threads are twisted together. 

10'.90s resultant counts. 

How many yards of single yarn will be required to 
make 8 lbs. of 2/ 20s worsted yarn ? 

89.600 yards. 

Find the equivalent counts of a 2 dram silk in 
denier counts. 34.66 denier or 34/36. 



CLOTp CRLCULflTIONS 



CLOTH CALCULATIONS, 



In dealing with cloth calculations, it may be best to 
explain the necessity for the rules that are given later. 

As is well known, cloth is made by interlacing warp 
and filling threads. When the threads cross each other 
they are bent more or less out of a straight line, a fact 
that causes the cloth to contract both in width and 
length. Thus the rate per cent of contraction will 
vary on different cloth constructions. 

For this reason, reeds of a lower count than the 
required count of the cloth must be used, making the 
cloth wider at the reed than on the roller. 

The length of a cut must be made longer on the 
slasher than the required length of the cut when 
woven. 

The threads of the warp must be sized before they 
can be woven, and this adds to the weight so that size 
and contraction are factors to be considered in all 
cloth calculations. 

The principal particulars of a plain cloth are the 
number of threads of warp and filling in one inch, the 
width of the cloth when woven, and the weight, which 
is designated as so many yards per pound. 

The number of warp threads per inch is known as 
the sley, and the filling threads as the picks, and when 
speaking of a cloth construction, the sley is invariably 
put first. A cloth that contains 56 warp threads and 
60 filling threads per inch, would be classed as a 
56 X 60 cloth. Warpthreads are commonly known 



48 

as ends, and this term will be used in all the calcu- 
lations. 

To find the number of ends required in a warp 
when the sley and width are g-iven: 

Rule. Mtiltiphj the sley hy the width and add extra 
ends for the selvages . 

Selvages are made by drawing some of the ends on 
the sides double, the number so doubled depending on 
the width of selvage desired. Extra ends must be 
added to keep the cloth at the required width. On 
heavy cloths, ply yarn is sometimes used for the 
selvage ends. 

Examples. 

IIow^ many ends are required to weave a cloth 56 
sley, 36 inches wide, with 24 ends extra for selvages? 

56 X 36 = '2016 + 24= 2040 ends. 

How many ends will be required to weave a cloth 
76 sley, 30 inches wide, allowing 36 ends for selvages ? 

76 X 30 = 2280 + 36 = 2316 ends. 

In the process of weaving, the warp and tilling 
threads are bent more or less out of a straight line, 
causing the' cloth to contract in both length and width. 
When the sley and picks are equal and the w^arp and 
tilling of the same counts, the contraction will be nearly 
equal in length and width. When the sley is higher 
than the picks the contraction in width will be less 
than if the picks were higher than the sley. If coarse 
tilling is used, the contraction in width will be less 
than if tine tilling is used, more especially if the coarse 
tilling is slack twisted. 

A plain cloth of ordinary construction about 50 sley 
and pick will contract about 7 per cent, while a cloth 



49 

100 sley and pick will contract about G per cent. 
Wlien warp stop motions are used, the warp has to be 
held at a greater tension on account of the drop wires 
and this extra tension causes the cloth to contract from 
1 to 2 per cent more, or from 8 to 9 per cent. 

There are two methods used for finding the dents 
per inch in a reed for a given sley cloth. One is to 
estimate how much a cloth will contract and make it 
that much wider in the reed, and from the sley re- 
quired and the estimated width at the reed find the 
dents per inch required in the reed. 

Another method is to make a calculation from the 
sley of the cloth required and use a rule that gives a 
sliding rate of contraction which decreases as the sley 
increases. 

By the first method, long practice and experience is 
absolutely necessary to estimate correctly the amount 
to allow for contraction, while the second method may 
be used by those less experienced. 

To illustrate the first method for finding the count 
of the reed, — suppose a cloth 100 sley, 40 inches wide 
is required, it would take 

100 X 40 = 4000 ends ^ 2 = 2000 dente. 

The next thing is to estimate the amount of con^ 
traction, which, under ordinary conditions, would be- 
about 2i inches, making the ends 42i inches in the 
reed 

•2000 dents ^ 42.50 inches = 47.05 dents per inch 
or a 47 dent reed. 

If this cloth were woven on a loom equipped with a 
warp stop motion, the width at the reed might be one 
inch more or 43^ inches, then, 

2000 dents ~ 43.50 = 45.97 dents per inch or a 
46 dent reed. 



50 

The width of cotton cloth is seldom changed in the 
process of finishing, while worsted and woolens which 
pass through fulling and washing processes may change 
in width considerably. Thus the finishing of the cloth, 
as well as the weave, must be taken into consideration 
in finding the reed for worsteds or woolen cloths. For 
these reasons, the above method is well suited to the 
woolen trade. 

The second method for finding the reed is to find 
the dents per inch from the sley of the cloth by rules 
that give a sliding rate per cent, of contraction which 
decreases as the sley increases. One of these rules is 
as follows : 

To find the dents per inch for any sley cloth: 

Rule. Subtract \ from the sley, from the result sub- 
tract b per cent. J divide this result by the Jiumber of ends 
per dent and aiiswer is dents per inch required. 

Examples. 

Find the dents per inch in a reed to weave a 48 
sley cloth, the ends to be drawn 2 in a dent. 
48 — 1 = 47 X .95 = 44.65. 
44.65 -f- 2 = 22.32 dents per inch required. 
To subtract 5 per cent., multiply by .95. 
If no contraction took place, a reed for a 48 sley 
cloth would require 24 dents per inch ; the rate per 
cent, for contraction allowed by the rule given is 7 per 
cent. 

24 — 22.32 = 1.68 dents less. 
1.68 ^ 24 = .07 or 7%. 

Find the dents per inch in a reed to weave a 96 
sley cloth, ends drawn 2 in a dent. 

96 — 1 = 95 X .95 = 90.25. 
90.25 -f- 2 = 45.12 dents per inch. 



51 

If no contraction took place, 48 dents would be 
necessary, but by the rule 45.12 dents, which is 6 per 
cent, less, is required. 

48 _ 45.12 == 2.88 dents less. 
2.88 -f- 48= .06 or 6%. 

The explanation of this rule is as follows : 
One deducted from any given number gives a higher 
per cent, reduction than the same amount taken from 
any higher number. This is where the varying rate of 
contraction is obtained by the rule. Five per cent, is 
then deducted in every instance, as per rule. 

To find the sley cloth a reed will weave when 
length and total dents in the reed are known : 

Rule. Divide the total dents by the length of the reed 
inside the reed bars ; the result is dents per inch; multiply 
dents per inch by ends per dent ; divide the result by .95 
and to this result add 1. 

Examples. 

What sley cloth will a reed weave that contains 
1030 dents on 36 inches, the ends drawn 2 in a dent? 
1080 -f- 36 = 30 dents per inch. 

30 X 2 = 60 ends per inch in reed. 
60 ^ .95 =: 63.15 + 1 = 64 sley cloth. 

If a reed contains 23.12 dents per inch, what sley 
cloth will it weave if the ends are drawn 3 in a dent ? 
23.12 X 3 = 69.36 ends per inch in the reed. 
69.36 -^ .95 = 73.01 + 1 = 74 sley cloth. 
Bars a quarter of an inch wide are put in the 
ends ot the reed to protect them from damage ; they 
also serve to have marked on them the number of 
dents, the length of the reed, and the sley of the cloth 
it would weave with the ends drawn 2 in a dent. 



52 

Looms are made in different widths, and are gen- 
erally known by the Ml width of cloth that can be 
woven on them. A loom that will weave cloth 40 
inches wide is styled a 40-inch loom, while a loom that 
will weave a cloth only 30 inches wide is styled a 30- 
inch loom, etc. 

Allowance is made for contraction in the width of 
the cloth, so that a 40-inch loom is made with about 
44 inches of reed space, while a 30-inch loom would 
hold a reed about 34 inches long. 

Sometimes the orders for narrow cloths exceed the 
possible production of the looms of the correct width, 
so that it is necessary to use the wider looms to 
weave it. 

When a cloth is woven that requires the full width 
of the loom, the reed is made to fill the reed space on 
the loom. When narrow cloth is woven on a wide 
loom, the reed is usually made about an inch longer than 
is actually required for the warp ends, and the reed 
space on the loom filled in with short pieces of old 
reeds. 

Selvage ends sometimes spring the dents in the reed, 
causing the ends to break. If some extra length is 
allowed on the reed, the selvage ends may be drawn in 
some other dents when a new warp is drawn, whereas, 
if the reed is only the exact length a new reed would 
be required. 

A method for finding the number of dents and the 
length of the reed, also the number ot harness eyes on 
each harness shaft will be explained by an example. 

Examples. 

Find the number of dents, the length ot the reed, and 
the number of eyes per shade on the harness to weave 
a 64 sley cloth in a loom, with 40 inches reed space. 



53 

64 — 1 = 63 X .95 = 59.85 4- 2 = 29.92 dents 
per inch in the reed and 29.92 eyes per inch on each 
shade of the harness, 2 shades per set. 

40 inches reed space less a half inch for reed bars 
equals 39.50 inches. 

39.50 X 29.92 = 1171 dents on 39.50 inches and 
1171 eyes on each shade of the harness : the reed will 
measure 40 inches with the reed bars. 

If the harness is made double shade, the number of 
eyes per shade would be one-half the number of dents 
in the reed. 

Find the number of dents in the reed, and harness 
eyes required on each shaft to weave a 5 end warp 
sateen, 114 sley, 40 inches wide, allowing a half inch 
extra width on harness and reed. 

114 X 40 = 4560 ends -f- 5 ends per dent =912 
dents required. 

114— 1 = 113 X .95= 107.35 -f- 5 = 21.47 dents 
per inch in the reed. 

912 ^ 21.47 = 42.47 inches space that 912 dents 
will occupy in the reed. 

42.47 4- -50 for extra width = 42.97 inches length 
of reed inside reed bars. 42.97 X 21.47 = 923 dents 
on 42.97 inches; add a half inch for reed bars and the 
reed will measure 43.47 inches over all. The harness 
will require 923 eyes per shade on 42.97 inches, 5 
shades in the set. 

When making a calculation to find the weight of 
filling in a given length of cloth, the width of the 
warp ends in the reed must be taken as the length of 
the filling pick, and not the actual width of the cloth. 

To find the width of the warp ends in the reed: 
Rule. Divide the total number of ends in the warp by 
the number of ends in a dent, equals the dents required if 



54 

all the ends were drawn 2 in a dent; subtract as many 
dents as are to contain double ends in the selvages, equals 
dents that actually contain ends ; divide by the dents -per 
inch in the reed, ecpials the width at the reed. 

Example. Find the width in the reed of a 64 sley 
cloth, 36 inches wide when woven, the ends to be 
drawn 2 in a dent with 48 extra ends in 12 dents for 
selvages. 

64 X 36 + 48 = 2352 ends. 

2352 ^ 2 == 1176 dents required if all dents 
contained 2 ends. 

48 of the 2352 are to be used 4 in a dent, thus there 
will be 1176 — 12 =1164 dents that contain warp 
ends. 

64 — 1 = 63 X .95 = 59.85 ^ 2 = 29.92 dents 
per inch in a 64 sley reed. 

1164 dents used -^ 29.92 = 38.90 inches wide at 
the reed. 

To find the weight of filling in a given length of 
cloth : 

Rule. Multiply the width at the reed by the picks per 
inch, result equals inches of filling in 1 inch of cloth, it 
also equals the yards of filling in 1 yard of cloth ; multi- 
ptly by length of cloth in yards, equals total yards of 
filling required; divide by 840. equals hanJcs; divide hanks 
by counts, equals weight in lbs. 

(The width at the reed multiplied by the picks per 
inch equals the inches of filling in 1 inch of cloth, 
which, divided by 36, equals the yards of filling in 1 
inch of the woven cloth. To obtain the yards of filling 
in 1 yard of cloth, multiply by 36 and the result is the 
same in yards of tilling in 1 yard of cloth as the inches 
in 1 inch of cloth, thus they will cancel and are omitted 
in the rule.) 



55 

Examples. 

Find the weight of filling in 50 yards of cloth woven 
with 72 picks of 60s filling and 32.50 inches wide at 
the reed. 

32.50 X 12 = 2340 inches of filling in 1 inch of 
cloth, also the yards in 1 yard of cloth. 

2340 X 50 = 117000 yards of filling in 50 yards 
of cloth. 

117000 ^ 840 = 139.28 hanks. 

139.28 -^ 60s = 2.32 lbs. of filling. 

If a cotton cloth is woven 64 X 68 — 36 inches 
wide, with 48 ends extra in 12 dents for selvages, what 
weight of 45s filling would be required to weave 100 
yards of cloth ? 

64 X 36 + 48 = 2352 ends. 

2352 -^ 2 =: 1176 — 12 = 1164 dents used. 

64 — 1 = 63 X .95 := 59.85 -^ 2 = 29.92 dents 
per inch. 

1164 -^ 29.92 = 38.90 inches wide at reed or 
length of each pick of filling, 

38.90 X 68 X 100 = 264520 yards of filling in 
100 yards of cloth. 

264520 -^ 840 = 314.90 hanks. 

314.90 -f- 45s filling = 6.99 lbs. filling req. 

To find the weight of filling of each color in a 
checked gingham when the picks of each color in the 
pattern, picks per inch, width at reed, and counts 
of filling are known : 

Rule. Multiply the width at reed by the picks per 
inch and this result by the cloth length, result equals total 
yards of filling required ; divide by 840, equals hanks ; 
divide hanks by counts, equals lbs. Divide the picks of 
each color in the pattern by the total pi^ks per pattern and 



56 

result equals the per cent, of each color of filling ; multi- 
ply the total weight of filling in the cloth length bij the per 
cent, of each color , ecpials the weight required of each color. 

Example. If a checked gingham is woven 30 inches 
at the reed, with 68 picks of 35s filling, what weight 
of each color of filling will it require to weave 100 
yards of cloth, with the following pattern ? 

40 picks white 

12 " blue 

8 " black 

6 " yellow 

66 " total picks in the pattern. 

40 -^ 66 = .61 or 61% white filling. 

12 -^ 66 = .18 " 18 '' blue " 

8 -^ 66 = .12 ^^ 12 " black " 

6 -^ 66 == .09 " 9 " yellow " 

30'/ X 68 X 100 ^ ^^ „ ^ . , 
=6.93 lbs. total weidit. 

840 X 35 "" 

6.93 X 61% = 4.23 lbs. white. 

6.93 X 18^' =: 1.25 " blue. 

6.93 X 12 " = 0.83 " black. 

6.93 X 9 " = 0.62 " yellow. 
6^93 " 

To find the weight of filling required for a given 
length of cloth when two different counts are used : 

Rule. Multij)ly the width at the reed by the picks per 
iiich and the result by the cloth length, result equals the 
total yards of fdliiig required ; divide by 840, equals total 
hanks. Divide the number of picks of each counts of 
filling in one repeat of the pattern by the total picks per 
pattern , result equals the per cent, of length of each counts. 
Multiply the total hanks by the per cent, equals hayiks of 
each counts, divide hanks by counts equals the required 
weight of each counts. 



57 

Example. Find the weight of each counts of filling 
required to weave 100 yards of cloth (filling welt 
pattern) woven 30 inches at the reed with 12 picks of 
60s filling for face, 4 picks of 10s filling for wadding 
per pattern, and 124 picks per inch. 

30^^X 124 X 100 ..^o^ , , ,, , 

=: 442.85 total hanks. 

840 

12 picks face -^16 picks per patt. =.75 or 75% in 
length of the 60s. 

4 picks wadding -f- 16 picks per patt. =.25 or 25% 
in length of the 10s. 

442.85 hanks X 75% = 332.13 hanks of 60s. 

332.13 ^ 60 = 5.53 lbs. of 60s. 

442.85 hanks X 25% = 110.71 hanks of 10s. 

110.71 -^ 10 = 11.07 lbs. of 10s. 

WARP CONTRACTION. 

The contraction that takes place in the length of the 
warp is caused by its being bent out of a straight line 
in passing around the filling. 

It will be plain that as the picks per inch increase, 
the rate of contraction will also increase, and that 
coarse filling will require a longer warp length than 
fine filling, because the warp ends will be bent more 
out of a straight line in passing around it. The sley 
will also have an effect on the rate of contraction, as 
for instance, a warp satin stripe with a plain ground 
may be woven from one beam, because the warp ends 
in the stripe are drawn four to six in a dent, and on 
account of their crowded condition are prevented from 
laying as flat as they do in a plain weave. 

The weave, sley, picks, counts of warp and filling 
and twist of yarns each have an effect on the con- 
traction of the cloth, and on account of the varied 



58 

constructions used in the manufacture of cotton cloths 
it is impossible to give a rule for finding the per cent, 
of contraction that will apply to all cloths. Long 
practice and familiarity with different cloths, and 
comparison of the cloth to be made with others of a 
smilar construction are the surest means for obtaining 
the desired results. 

For plain cloths of ordinary construction the 
following rule may be used to find the approximate 
per cent, of contraction from warp to cloth : 

Rule. Multiply the picks per inch by 3.5 and divide 
the result by the counts of the filling. 

Example. Find the slasher length of a cut with 68 
picks of 32s filling, the cut to be 50 yards when woven. 

68 X 3.5 = 238. 

238 -f- 32 = 7.4 per cent contraction. 

50 X T.4% = 3.70 yards. 

50 + 3.70 = 53.70 " slasher length. 

When a cloth is made where the construction is 
such that the contraction would be more than ordinary, 
4 may be used as a multiplier, or if there is reason to 
expect a less rate of contraction, 3 may be used, etc. 

On fancy cloths, yarns of different counts are often 
used ; ply yarns are also used either for backing, as in 
bedford cords, or for figured work, as on lenos and 
lappets. 

When making new patterns. on such cloths, there is 
no rule by which the length of yarn required to weave 
a given length of cloth can be obtained, so that samples 
must be woven and the lengths and weights obtained 
in that way. 

When it is desired to reproduce a cloth from a small 
sample, a piece from three to six inches, more or less, 



59 

as may be obtainable, is taken and a warp end removed 
from each distinct weave, straiglitened out, and 
compared with the length of the cloth ; from the differ- 
ence in the lengths of the end and the cloth the per 
cent to allow for contraction on the cloth length may 
be obtained. 

To find the per cent to allow for contraction from 
a cloth sample : 

Rule. Pull a warp end out of the cloth sample^ 
straighten it out and measure it ; divide the difference in 
the lengths of the end and the cloth by the length of the 
cloth, the result equals the per cent, to allow for contraction 
on the cloth length. 

Examples. 

If an end from a piece of cloth 3 inches long 
measures 3.25 inches, what is the per cent, to allow for 
contraction on the cloth ? 

3.25 — 3 == .25 difference in length of the end and 
the cloth. 

.25 ^ 3 = .083 or 8.3% contraction. 

If an end from a piece of leno cloth 4 inches long, 
measures 6 inches, what is the per cent, of contraction 
to allow on the cloth ? 

6 — 4^2 inches difference. 

2 ^ 4 = .50 or 50% contraction. 

It will be clear that if 4 inches of cloth require 6 
inches of yarn, the difference in the lengths of the yarn 
and the cloth is half the length of the cloth or 50% . 

If an end from a piece of cloth 4 inches long, measures 
4.50 inches, what length of yai-n must be run on the 
slasher to obtain a cut of cloth 60 yards long ? 



60 

4.50 — 4 = .50 difference in length of the end and 
the cloth. 

.50 ^ 4 = .125 or 12.5% contraction. 

60 yards X 12.5% = 7.5 yards. 

60 -\- 7.5 = 67.5 yards of yarn required on slasher 
to weave a 60 yard cut. 

To find weight of warp yarn in a given length of 
cloth : 

Rule. Mnltiply the ends of each counts by the slasher 
le7igth, result equals yards of ivarj); divide by 840, equals 
hanks; divide hanks by warp counts, equals weight. If 
the tveight when sized is desired, add from 5 to 8%, ac- 
cording to the amount of size to be put 07i the yarn. Ply 
yarns, except when mercerized, do not require size. 
Examples. 

If a cotton cloth is constructed 68 sley, 36 inches 
wide, with 40 ends extra added for selvages, what 
weight of 36s warp yarn would be contained in 50 
yards of cloth, allowing 5 per cent, for contraction and 
7 per cent, for size ? 

68 X 36 + 40 == 2488 ends. 

50 rds X 5% = 2.50 yards. 

50 " -{- 2.50 = 52.50 yards slasher length. 

2488 X 52.50 = 130620 total yards warp. 

130620 -^ 840 = 155.50 hanks. 

155.50 -^ 36s = 4.31 lbs. without size. 

4.31 lbs. X 7% size = .30 lbs. 

4.31 -\- .30 = 4.61 lbs. weight of warp with size 
added. 

When the cloth contains ply yarns, the counts of the 
single yarn is considered; as for example, 100 ends ot 
2/40s would be calculated as 200 ends of 40s. 

When the ends in one part of the cloth vary from 
another part, either in counts or in the rate of con- 
traction, they must be put on separate beams. 



61 

If a cloth is woven witli 2100 ends of 40s with 5 
per cent, contraction, and 7 per cent, size, 200 ends of 
2/ 15s, with 15 per cent, contraction, and 350 ends of 
3/20s, with 25 per cent, contraction, what weight of 
warp yarn will be required to weave 100 yards of 
cloth ? 

2100 X 105 = 220500 yards -^- 840 = 262.50 
lianks. 

262.50 -^ 40s. = 6.56 lbs. + 7% size = 7.01 
lbs. weight of the 40s. yarn. 

200 X 2 X 115 = 46000 yards -^ 840 = 54.76 
hanks. 

54.76 -^ 15s = 3.65 lbs. of 15s yarn. 

350 X 3 X 125 = 131250 yards -^ 840 = 156.25 
hanks. 

156.25 -^ 20s = 7.81 lbs. of 20s yarn. 

7.01 -f 3.65 H- 7.81 = 18.47 lbs. total weight. 

AVERAGE NUMBERS. 

One of the principal particulars in a plain cloth is 
the weight, which is designated in yards per lb., with 
a given sley, picks, and width. 

When the numbers of the warp and filling are not 
given, they may be found by first finding the average 
numbers, i. e., a number, which, if used for warp, and 
filling, would give the weight desired ; then to assume 
a number to be used for the warp that is coarser than 
the average numbers in the cloth and one that is 
perhaps already being produced, then by a calculation 
find the numbers of the filling required to give the 
yards per lb. 

On the coarser grades of cloth, the difi*erencc in the 
numbers of the warp and filling will be slight, but on 
the finer grades the difference will be greater. 



62 

The reason for this is that warp yarns are subjected 
to greater strain than filling yarns and a stronger yarn 
is necessary. 

To find average numbers when sley, picks, width, 
and yards per lb. are known : 

Rule, First find number of eiids, then multiply by 
estimated slasher length, result equals yards of tear p ; 
divide by 840, equals hanks; add a certain per cent to 
represent size, and consider it as yarn. Multiply ividth 
at reed, by picks per inch, result equals iiiches of filling in 
one inch of cloth, also the yards of filling in one yard of 
cloth ; midtiply by length of cloth, result equals yards of 
filling; divide by 840, equals hanks. Divide length of 
cloth by yards per lb., resiilt equals weight of cloth. Add 
hanks of warp and filling and divide by weight of cloth, 
equals average numbers. 

In cloth calculatiolis where the length is not specified, 
100 yards will be found the most convenient length to 
use on account of the percentages for contraction and 
from the fact that it is an easier sum to use than any 
part of 100. 

Example. Find average numbers on a cloth 48x 52, 
36 inches wide, and weighing 6 yards per lb. 

48 X 36 + 40 == 1768 ends. 

1768 X 106 yards (6% added for contraction) = 
187408 yards. 

187408 ^ 840 = 223.10 hanks. 

223.10 + 7% for size = 238.71 hanks as re- 
presented with the addition of size. 

1768 -^ 2 = 884—12 = 872 dents occupied with 
yarn. 

48 — 1 = 47 X .95 = 44.65 -i- 2 = 22.32 dents 
per inch. 



63 

872 total dents ^ 22.32 = 39.06 inclies wide at 
reed. 

39.06 X 52 X 100 = 203112 yards of filling in 
100 yards of cloth. 

203112 4- 840 — 241.80 hanks filling. 

238.71 hanks warp + 241.80 hanks filling=:480.51 
total hanks. 

100 yards of cloth -^ 6 yards per lb. = 16.66 lbs. 
of cloth. 

480.51 4- 16.66 = 28.84 average numbers. 

It will be seen from this problem, as in yarn calcu- 
lations, that having found the amount of yarn in hanks 
it is divided by the weight to find the counts. 

To find the average numbers when sley, picks, 
and counts of warp and filling are known: 

Rule. Divide the sley by the icarp counts, result equals 
relative weight of ivarj) yarn ; divide the picks by the 
filling counts, ecpials relative iveight of filling yarn; 
divide the sum of the sley and picks by the sum of the 
relative weights, e(pials the average counts. 

Example. If a cloth is woven 68 X 72 with 34s 
warp and 50s filling, what is the average numbers ? 
68 ^ 34 = 2 relative weight of warp. 
72 -f- 50 = 1.44 ^' """ " filling. 

140 3^ 

140, the sum of the quantities -^ 3.44, the sum of 
the relative weights = 40.69 average numbers ; or, 

The sley and picks, being quantity of yarn in a given 
space, it may be considered as hanks, and 
68 hanks of warp 4- 34s counts = 2 lbs. of warp. 
72 '^ ^' filling -^- 50s " =1.44" "filling. 

140 " 3.44 " total. 

140 hanks -^ 3.44 lbs. = 40.69 average counts. 



64 

On fancy cloths, the rate of contraction on the 
different weaves that are combined in the cloth, is often 
unequal, which makes it necessary to use more than 
one loom beam. Sometimes ply yarns are used, and in 
this case, it will require separate beams ; therefore it 
follows that the different yarns must be calculated 
separately. 

To find average numbers when the cloth contains 
more than one counts of warp, and the number of 
ends of each counts is known, with picks per inch, 
filling counts, and width of cloth at reed : 

Rule. Multiply each member of ends by the slasher 
length, result equals yards of warp ; divide length of 
ivarji by 840, equals hanks ; divide hanks by counts, 
equals iveight. Multiply width at reed, by picks per inch, 
equals yards of filling i7i one yard of cloth; imiltiply by 
length of cloth or cut length, equals total yards of filling; 
divide by 840, equals hanks ; divide hanks by counts, 
equals iveight. Add all the weights and divide into the 
total number of hanks of warp and, filling and result is 
average number. 

Example. If a cloth is 30 inches wide at reed and 
contains 76 picks per inch of 70s filling and requires 
the following number of ends, what is the average 
number ? 

2000 ends 60s with 4% contraction. 
300 " 2/15s " 15 '^ ^' 

600 " 3/20s " 25 " " 

^^^iAj^* = 247.61 hanks- 60s = 4.12 lbs. 

840 

30^2^^)015^ 82.14 " ^15s= 5.42 " 
840 



65 
600 X 3 X 125 



267.85 '^ -^ 20s =13.36 lbs. 
271.42 '• -f-70s= 3.87 " 



840 
30^^ X 76 X 100 

^^^ 869.02 '^ 26.77 " 

869.02 -^ 26.77 = 32.46 average numbers. 

To find the average number when the number of 
ends and picks per inch, width at reed, and yards 
per lb. are known : 

Rule. Maltiply each numher of ends by its ow7i 
slasher length, add a given per cent, for size to the single 
yarn, and consider the increase as yards ; multiply the 
width at the reed by picks per inch, equals the yards oj 
filling in one yard, of cloth ; multiply by length of cloth, 
equals total yards ofjilling ; add the yards of warp and 
filling and divide by 840, equals hanks of yarn ; divide 
hanks by weight in lbs., equals counts, which, in this case, 
will be the average counts of the warp and filling. 

The length of the cloth, on which the calculation is 
made, divided by the yards per lb., equals the weight 
of the cloth. 

Example. If a cloth contains 2200 ends of 50s 
which contract 5 % and carry 5 % size, 400 ends of 
2/30s which contract 25%, 200 ends of 3/15s which 
contract 20%, and 76 picks per inch, is 34 inches at 
the reed, and weighs 5 yards per lb., what is the 
average counts of the warp and filling ? 

As the single yarn contracts 5% and carries 5% 
size it equals 10% of an increase in its weight, and if 
the calculation is made on 100 yards of cloth it equals 
10 yards; — thus, 



66 

2200 X 110 = 242000 yards of 50s 
400 X 2 X 125 = 100000 " " 30s 
200 X 3 X 120 = 72000 ^' '' 15s 
34^' X 76 X 100 = 258400 " " filling. 

672400 yards of yarn. 
672400 -f- 840 = 800.47 hanks. 
100 yards of cloth -i- 5 yards per lb. = 20 lbs. of 
cloth. 

800.47 hanks -f- 20 lbs. = 40.02 average counts. 

It has been explained how the warp and filling yarns 
contract in the process of weaving, and that the con- 
traction and the size which is added to the single warp 
yarn increase the weight. 

In making a calculation on cloth, contraction is 
considered by adding to the cloth length an amount 
which varies on difi'erent cloth constructions ; cloth 
made from coarse yarns, however, will contract more, 
and will also absorb more size than finer yarns. 

If a thread is removed from a piece of woven cloth 
it will be found to have a wavy appearance, caused by 
its bending around other threads; consequently, in 
this condition, it will weigh more than a thread that is 
perfectly straight. 

The standard lengtli of the cotton hank is 840 yards, 
but if this length is put into cloth it reduces its length 
which will vary according to the construction of the 
cloth. 

In previous examples, the contraction has been 
added to the cloth length or width and the warp and 
filling calculated separately. 

Shorter methods are obtained by adding the amount 
of warp and filling as it appears in the cloth and 
making the allowance for contraction and size on the 
yards in the hank. 



67 

A list of lengths or constants is given that may be 
used in place of 840 yards, also the yarns on which 
they may be used. It will necessarily require some 
experience to determine the most suitable number to 
use on any given class of goods, but when once ob- 
tained, their use will be found to be practical. 
Use 745 yards on counts from 18 to 24 
'^ 750 ' ^' ^' '' '^ 25 '^ 35 

u 755 u u u u 36 u 45 

u 760 u u u u 46 u 60 

" 765 " " " ^^61 '^ 80 

u 770 '.' u u u gi i, 100 

The counts of yarns given are the average of warp 
and filling. 

To find average number on plain cloths when 
sley, picks, width and yards per lb. are known : 

Rule. Add sley and jjicks, result equals inches of 
yarn in one inch of cloth ; multiply by the width, equals 
inches of yarn in one inch of cloth of that width; it also 
equals the yards of yarn in one yard of cloth; multiply 
by yards per lb., equals the yards of yarn in one lb. of the 
cloth ; divide by the most suitable constant or length in 
place of 840, equals counts, which will be the average of 
waip and, filling. 

Examples. 

If a cloth 52 sley, 48 picks, 36 inches wide, weighs 
4 yards per lb., what is the average numbers ? { Use 
745 for constant). 

(52 +48) X 36 X 4 



745 constant 



= 19.32 average numbers. 



If a cloth 84 X 88 — 32 inches wide, weighs 11 yards 
per lb., find the average numbers. (Use 760 for a 
constant). 



68 

(84+ 88) X 32 X 11 
760 constant 



= 79.66 average numbers. 



To find the average counts of warp yarns when 
two or more different counts are used : 

Rule. Divide each number of ends by its own counts, 
equals relative iveight ; add the weights and divide into 
the total number of ends. 

Examples. 

If a cloth contains 2800 ends of 70s and 400 ends of 
40s, what is the average counts of the warp ? 
2800 -f- 70 = 40 relative weight. 
400 -^ 40 = 10 '^ 



3200 total ends 50 total 
3200 -f- 50 = 64 average counts of warp. 

If a cloth contains 2200 ends of 60s, 300 ends of 
3/ 15s and 200 ends of 2/40s, what are the average 
warp counts ? 



60 = 36.66 rel. wgt. 
15 = 60.00 
40 = 10.00 



2200 ends of 60s ^ 2200 
300 >^ 3/15s =: 900 
200 " 2/40s = 400 

3500 total 106.66 
3500 ^ 106.66 = 32.81 average warp counts. 

To find the yards per lb., when sley, picks, 
width and counts of warp and filling are known : 

Rule. Multiply the ends by slasher length, result 
equals yards of warp ; divide by 840, equals hanks ; di- 
vide hanks by warp counts, equals weight ; add a certain 
per cent, for size, equals total weight of warp yarn. Mul- 
tipty width at the reed by picks per inch, equals yards of 
filling in one yard of cloth ; multiply by cloth length, 
equals total yards of filling ; divide by 840, equals hanks; 



69 

divide hanks hy JiUlncr roiints, rqimls ireight of filling 
xjarn ; add weight of warp and filling and divide into 
cloth length, equals yards per lb. 

Examples. 

Find the yards per 11). of a cloth with the following 
particulars : . 

56 X 60 — 36 inches wide, 20s warp and 24s filling. 
60 X 3.5 = 210 -^ 24 = 8.75% for warp con- 
traction. 

56 X 36 + 40 for selvages = 2056 ends. 

2056 X 108.75 yards slasher len,o:th =: 223590 
yards of warp. 

223590 -f- 840 = 266.17 hanks. 
266.17 ~ 20s = 13.35 lbs. + 7% size = 14.2S 
lbs. weight of warp sized. 

2056 -f- 2 =: 1028—12 ~ 1016 dents that contain 
yarn. 

56—1 = ^5 X .95 = 52.25 ^ 2 =z 26.12 dents 
per inch. 

1016 -f- 26.12 =: 38.89 inches wide at reed. 
38.89 X 60 X 100 = 233340 yards of filling in 
100 yards of cloth. 

233340 ^ 840=: 277.78 hanks. 

277.78 -^ 24s = 11.57 lbs. filling. 

14.28 + 11.57 = 25.85 lbs. warp and filling. 

100 yards -^ 25.85 = 3.86 yards per lb. 

If a piece of cloth contains 1800 ends of 40s, with 
5% contraction and 6% size, 200 ends 2/35s with 
10% contraction, 500 ends 3/45s with 17% contrac- 
tion, 72 picks of 60s filling, 30 inches wide at reed, 
what are the yards per lb. ? 



70 
5.62 lbs. of yarn+6% size=5.95 lbs. 
= 1.49 " 

= 4.64 ^' 

=z 4.28 " 



1800 X 105 
840 X 40 
200 X 2 X 110 _ 

840 X 35 ~ 
500 X 3 X 117 _ 

840 X 45 ~ 
30^^X 72 X 100 _ 

840 X 60 ~~ 

Total weidit, 16.36 lbs. 

100 yards of cloth -^ 16.36 lbs. = 6.11 yds. per lb. 

By making use of the constants as explained under 
average numbers, a shorter method for finding the 
yards per lb. is obtained. 

To find the yards per lb. when the sley, picks, 
width, and counts of warp and filling are known : 

Rule. First find the average numbers by short method 
as previously explained, then add the sley and picks, re- 
sult e(pials inches of yarn in one square inch of cloth ; mul- 
tiply by width of the cloth, equals inches of yarn in the 
cloth width and one inch long ; it also equals the yards of 
yarn in one yard, of the cloth. Multiply the average num- 
bers by the most suitable constant, equals yards of yarn in 
one lb. of that average number with allowance made for 
contraction and, size. Divide the yards of yarn in one lb. 
of the average number by the yards of yarn in one yard 
of the cloth, equals the yards of cloth in one lb. 

Example. If a cloth is woven 56x60, with 20s 
warp and 24s filling, and 36 inches wide, what are the 
yards per lb. ? 

56 -f- 20 = 2.80 relative weight of warp. 

60 -^ 24 == 2.50 " '' ^' filling. 

116 5.30 total weight of yarns. 



71 

116 ^ 5.30 = 21.88 average numbers. 
21.88 X 745 = 16300 yards in 1 lb. with allow- 
ance for contraction and size. 

(56 + 60) X 36 = 4176 inches of warp and filling 
in a strip of cloth one inch long, 36 inches wide ; it 
also equals the yards of yarn in one full yard of the 
cloth. 

16300 ^ 4176 = 3.90 yards of cloth in one lb. 
or 

21.88 X 745 ,^^ , 

= 3.90 yards per lb. 

(56 + 60) X 36 

To find the yards per lb. from short lengths of 
cloth : 

Rule: Multiply 7000 grains by the number of square 
inches of cloth weighed, divide the result by the weight of 
the cloth in grains multiplied by the width of cloth desired, 
and by SQ inches per yard. 

Example. If a piece of cloth 3 inches square 
weighs 1 1 grains, what would be the yards per lb. if it 
were woven 30 inches wide ? 

3 X 3 X 7000 .oc, A 11 

=5.30 yards per lb. 

11 X 30 X 36 

The problem is worked by proportion. The square 
inches in the piece of cloth weighed is to its weight as 
the square inches in one yard is to the grains in one lb. 

30 X 36 ^ 1080 square inches of cloth in one 
yard of cloth 30 inches wide ; then if 9 square inches 
weigh 11 grains, 1080 square inches will weigh 

1080 X 11 
= 1320 grains, weight of 1 yard. 

V 

7000 grains in one lb. -h 1320 grains, weight of one 
yard =:r 5.30 yards per lb. 



72 

More accurate results are obtained by weighing 
larger pieces, and when possible a full yard should be 
weighed and its weight divided into 7000 grains. If 
any fractional part of a yard is weighed, its weight in 
grains must be divided into the same fractional part of 
7000. 

Examples. 

If a yard of cloth weighs 650 grains, how many yards 
will weigh 1 lb. ? 

7000 ^ 650 = 10.85 yards per lb. 

If a quarter yard of cloth should weigh 250 grains, 
how many yards will weigh 1 lb. ? 

7000 -^ 4 = 1750 grains in a quarter lb. 
1750 -f- 250 = 7 yards per lb. 

In mills where a large variety of cloths are made, it 
is the practice to spin but three or four different counts 
of warp yarn, and to regulate the weights of the cloth 
by varying the counts of the filling. 

It is economy to manufacture cloth l)y this method, 
on account of the various processes warp yarn must 
pass through, while filling may be taken to the loom 
direct from the spinning frame or mule. 

To find the counts of filling required to govern 
the yards per lb., when sley, picks, counts of warp, 
width and yards per lb: are known : 

Rule. MnltipJy ends by slasher length, equals yards 
of warp ; divide by 840, equals hanlts ; divide hanks by 
counts, equals weight ; add a certain per cent, for size 
ivhen required J equals the weight of the warj) ivith size 
added. When more than one counts of warj) yarfi are 
2ised in the cloth each count must be calculated, separately 
and their weights added for the total weight of . the warp 
yarn. Multiply luidth at the reed by jncks per inch, 



73 

apials ijards ofJiUing in one yard of cloth ; midtiphj by 
doth length, equals total yanh of Jilling ; divide by 840, 
equals hanks of filling. Divide length of cloth by yards 
yer lb., equals weight of the cloth. Subtract iveight of 
warj), and the balance is the weight of the filling ; divide 
the hanks of Jilling by the weight of filling, equals the 
counts of filling required. 

Examples. 

Finds the counts of filling required on a cloth made 
with the following particulars : 

88 X 92 with 60s warp, 32 inches wide and 11 yards 
per lb ; 4% warp contraction and 6 % size. 

88 X 32 + 40 = 2856 ends. 

2856 X 104 = 297024 yards of warp. 

297024 -f 840 = 353.60 hanks ^ 60 = 5.89 lbs. 

5.89 + 6% for size = 6.24 lbs. weight of warp 
with size added. 

100 yards of cloth ^ 11 yards per lb. — 9.09 lbs. 
of cloth. 

9.09 — 6.24 lbs. of warp = 2.85 lbs. of filling. 

2856 ^ 2 = 1428 — 12 = 1416 dents filled with 
yarn. 

88—1 == 87 X .95 = 82.65 ^ 2 — 41.32 dents 
per inch. 

1416 _i_ 41.32 = 34.26 inches wide at the reed. 

34.26 X 92 X 100 — 315192 yards of filling. 

315192 ^ 840 = 375.22 hanks of filling. 

375.22 -^ 2.85 lbs. = 131.65 counts of filling re- 
quired. 

If a cloth is woven 30 inches wide at reed, with 76 
picks per inch, and contains 2000 ends of 60s warp 
with 5% contraction and 6% size,. 300 ends of 2/15s 
with 15% contraction, 200 ends of 2 /40s with 25% 
contraction, the cloth to weigh 6.50 yards per lb., what 
counts of filling is required ? 



7.4 

2000 X 105 



= 4.16 lbs. warp + 6% size == 4.41 lbs. 

= 5.47 " 
= 1.48 " 



840X60 
300X2X115 

840X15 

200X2X125 

840X40 

Total lbs. warp 11.36 

30^^ X 76 X 100 

= 271.42 hanks filling. 

■840 * 

100 yards cloth -^ 6.50 yards per lb. = 15.38 lbs. 
of cloth. 

15.38 — 11.36 lbs. warp = 4.02 lbs. of tilling. 

271.42 hanks of filling -^ 4.02 lbs. = 67.51 counts 
of fillina: required. 

If the constants 745/770 are used, the process for 
finding the filling required is made considerably 
shorter. 

To find the filling required to govern the yards 
per lb. (on plain cloth) when the sley, picks, width 
and yards per lb. are known : 

Rule. First find the average numbers, (see short 
method for finding average numbers) the7i add the sley 
and picJcSy result equals quantity of yarn in one square 
inch ; divide by average numbers, equals a sum that re- 
'presents the weight of ivarj) and filling ; divide the sley 
by the counts of the ivarp, equals the relative weight of the 
warp ; deduct the relative weight of the warp from the 
total, the balance represents the relative weight of the 
filling; divide the picks by the iveight of the filling, 
equals the counts of filling required. 

Example. If a cloth is made 88 x 92, 32 inches 
wide, with 60s warp, and is required to weigh 11 yards 
per lb., what counts of filling are required ? 



75 

(88 + 92) X 32 X 11 ^ ^^.82 average number. 

765 
88 + 92 = 170 ^ 82.82 = 2.17 total weight. 
88 -^ 60s warp — 1.46 relative weight of warp. 
2.17 — 1.46 r= .71 relative wei^^ht of filling. 
92 -^ .71 = 129.57 counts of filling required. 

It has been previously explained that quantity 
divided by counts equals weight. In this case, sley + 
picks equals the quantity and may be considered as so 
many hanks; i.e., 88 hanks of warp and 92 hanks of 
filling, 88 + 92 = 170 total hanks. 

170 hanks ^ 82.82 counts = 2.17 lbs. 

88 hanks of warp -^ 60s counts = 1.46 lbs. of 
warp. 

2.17 — 1.46 — .71 lbs. of filling. 

92 hanks of filling ^ .71 lbs. = 129.57 counts of 
filling required. 

The approximate counts of warp and filling from 
a small sample of ordinary plain cloth may be found 
as follows : 

Example. If a piece of cloth 4x5 inches, weighs 
14 grains and counts 76 X 80, what counts of warp 
and filling could be used if it were woven 30 inches 
wide? 

First, find yards per lb. by short method. 

4 X 5 X 7000 ^ ^_ , ,, 
z= 9.26 yards per lb. 

14 X 30 X 36 
Second, find average number by short method. 

(76 + 80) X 30 X 9.26 ^^^^ , 

^- ^ i = 57.02 average number. 

760 
Use a 50 warp and find filling required. 



76 

76 + 80 ^ 57.02 = 2.73 total weight. 
76 sley ^ 50 warp = 1.52 weight of warp. 
2.73 — 1.52 = 1.21 weight of filling. 
80 picks -f- 1.21 weight of filling = 66.11 counts of 
filling. 

To find the per cent, of warp and filling in a piece 
of cloth, is to find the weights of the different yarns 
used in its construction and divide the weight of 
each by the weight of the whole. 

Example. If a cloth is constructed 88 X 92, 32 
inches wide, with 60 warp, and 125 filling, allowing 5 
per cent, for warp contraction and 5 per cent, for size, 
what is the per cent, of warp and filling ? 

Rules have been previously given for the following 
work : 

88 X 32 + 30 = 2846 ends. 

2846 X 105 

= 5.92 lbs. of warp + 5% ^o^* size = 

840 X 60 6 21 lbs. of warp and size. 

2846 -f- 2 = 1423 — 12 = 1411 dents required. 

88 — 1 =r 87 X .95 =r 82.65 ^ 2 = 41.32 dents 
per inch in the reed. 

1411 -^ 41.32 =z 34.14 inches wide at the reed. 

34.14 X 92 X 100 

r^ 2.99 lbs. of filling. 

840 X 125 

6.21 lbs. warp + 2.99 lbs. filling = 9.20 lbs. total. 

6.21 ^ 9.20 = .67 or 67% warp. 

100% — 67% =33% filling. 

A shorter method is to find the relative weights of 
warp and filling, and divide the relative weight of warp 
by the sum of the relative weights ; answer is the per 
cent, of warp ; then deduct the per cent, of warp from 
100%, result equals the per cent, of filling. 



77 

The last example worked by this method would be 
as follows : 

88 sley ^ 60 warp = 1.46 relative wgt. of warp. 
92 picks -f- 125 filling = ^ -' " " filling. 

2.19 total weight. 
1.46 -^ 2.19 = .66 or 66% warp. 
100% — 66% = 34% filling. 

To find per cent, of warp and filling when ends, 
picks, and counts of warp and filling are known : 

Rule. Divide ends by warp counts, result equals 
rclatlvx weight of warp ; multiply picks by ividth and 
divide result by Jilling counts, equals relative weight of 
filling ; divide iveight of warp by weight of warp and 
filling, equals per cent, of warp ; deduct per cent, of warp 
from 100%, equals per cent, of filling. 

Example. If a cloth 30 inches wide contains 2560 
ends of 50s warp and 72 picks per inch of 60s filling, 
what is the per cent, of warp and filling? 

2560^50 ^ 51.2 relative weight of warp. 

'^ ^^ ^^ = 36.0 - - " filling. 

bO 

87.2 weight of warp and filling. 

51.2 -^ 87.2 = .58 or 58% warp. 
100% —58% =42% filling. 

In one yard of the cloth in the last example, 2560 
ends equal the yards of warp yarn in one yard of cloth ; 
the picks per inch multiplied by the width equals the 
yards of filling in one yard of cloth; thus, if the amount 
of warp and the amount of filling is each divided by its 
own counts the result will equal the proportionate 
weight of each as per rule. 



78 

To find the per cent, of warp and filling in a cloth 
when sley, picks, average counts, and warp counts 
are known : 

Rule. Divide the sum of sley and jncJcs by average 
counts, ecpials weight of u^arp and Jilling ; divide sley by 
warp coujits, equals relative weight of warp; divide weight 
of warp by weight of warp and filling, ecpials per cent, of 
warp; deduct per cent, of warp from 100%, equals per 
cent, of filling. 

Example. If a cloth is made 8S X 84, with 45s 
warp and an average number of 50, what is the per 
cent, of warp and filling ? 

88 4- 84 

z= 3.44 weiorht of warp and fillinp^. 

50 ^ V ^ 

88 -f- 45 = 1.95 relative weight of warp. 
1.95 -^ 3.44 = .56 or 56% warp. 
100% — 56%= 44% filling. 

To find the per cent, of warp and filling when the 
cloth contains more than one counts of warp, and 
the counts of warp and filling and width at reed are 
known : 

Rule. Multiply each iiumber of ends by its slasher 
length ; divide by 840, equals hanks ; divide hanks by 
coimts, equals iveight ; add all the weights, equals weight 
of warp yarns. Multiply width at reed by picks per inch 
and restdt by cloth length, equals yards of filling ; divide 
by 840, equals hanks ; divide hanks by counts, equals 
weight of filling ; add weight ,of vmrp and filling and 
divide into the weight of ivarp, equals per cent, of warp ; 
deduct per cent, of warp from iOOfo , equals the per cent, 
of filling. 

Example. If a cloth is woven 30 inches at reed 
with 76 picks of 70s filling, and contains the following 
warp ends, what is the per cent, of warp and filling ? 



79 



2000 ends of 


60s 


witli 


5% 


contraction. 


300 


U ii 


2/15s 


a 


15% 


u 




200 


u u 


2/40s 


u 


25% 


u 




2000 


X 105 







4.16 lbs. 


of 60s. 



840 X 60 






300X 2 X 115 _ 
840 X 15 


5.47 


" " 15s. 


200 X 2 X 125 _ 
840 X 40 


1.48 
11.11 


" " 40s. 
^' warp. 


30^^X76X100 _ 


3.87 lbs. of 


70s fillins-. 



840 X 70 
11.11 + 3.87 = 14.98 lbs. of warp and filling. 
11.11 -^ 14.98 = .74 or 74% warp. 
100% — 74% =26% fillincr. 

When the per cent, of each counts of warp is 
required, divide^ the weight of each counts by the total 
weight of the warp and filling. 

In the last example there are 4.16 lbs. of 60s; 5.47 
lbs, of 15s; 1.48 lbs. of 40s; and 3.87 lbs. of filling, 
making a total of 14.98 lbs. of warp and filling. What 
is the per cent, of each counts of warp ? 



14.98 = .28 or 28% of 60s warp. 
14.98 = .36 or 36% " 15s " 



4.16 
5.47 
1.48 -^ 14.98 = .099 or 9.9%" 40s '^ 

73.9% warp. 
100% — 73.9% = 26.1% filling. 

On many fancy cloths, the ends are not drawn in the 
reed in regular order, and some dents contain more 
ends than others ; therefore other methods than those 
previously given must be used to find the number of 
ends in a warp, also the count of the reed to use, etc. 



80 

On plain cloths, tlie sley means the number of ends 
in each inch of cloth ; on fancy cloth, the sley is under- 
stood to be the average number of ends in one inch, 
which is found by dividing the total ends by the width 
of the cloth. 

Example. If a cloth, 30 inches wide, contains 2000 

ends of 60s, 250 ends of 2/ 15s, and 100 ends of 

3/ 20s, what is the average sley of the cloth ? 

2000 X I = 2000 ends of 60s. 

250 X 2 =: 500 '^ " 15s. 

100 X 3 rr= 300 " " 20s. 

2800 total ends. 
2800 ~ 30^^ = 93 ends per inch or average sley. 

If this cloth were woven with an 80 sley reed and 84 
pick per inch, the particulars would be specified as 
follows : 

80 X 84 

93 
showing that an 80 sley reed is used and that the cloth 
averages 93 sley and 84 picks per inch. 

The term sley reed is understood to be a reed that 
will weave a given sley cloth with the ends drawn two 
in each dent, but when more or less ends are drawn in 
a dent and every dent does not contain the same 
number, the sley given is the average number of ends 
per inch. 

To find the number of ends in a cloth with an ir- 
regular reed draft, when the ends in one pattern, the 
sley reed, and the width of the cloth is given: 

Rule. Divide the sley of the reed by tiro, equals the 
dents per inch in the cloth after contraction ; multiply by 
the width of cloth required, equals total dents required ; 
deduct as many dents as may be required for selvageSf 



81 

equals dents to be used for the body of the cloth; divide by 
the dents in one pattern^ equals the number of patterns ; 
multiply the ends in one iiattern by the number of patterns, 
equals ends required ; add extra, ends to he used, for 
selvages, equals total ends. 

Example. If a cloth is made with the following 
reed draft on a 76 sley reed, how many ends are 
required to make it 28 inches wide? 

Dents. Ends. 

14 28 

5 20 

6 12 
J_ 20 

30 80 

76 -^ 2 = 38 dents per inch as represented in the 
cloth after contraction. 

38X28^^ = 1064 — 12 for selvages = 1052 dents 
required for the body of the cloth. 

1052 -^ 30 dents per patt ==35 patts and 2 dents 
over. 

If this warp were drawn as per draft, the cloth 
would contain 14 dents, 2 ends in a dent, next to the 
selvage on one side, and 5 dents, 4 ends in a dent, on 
the other. This would not look just right and it is 
the practice in designing a cloth to have a few dents 
plain weave next to the selvages when it is possible. 
In this case, the 2 dents over the 35 patterns would be 
utilized by adding them to the 14 dents plain, making 
16 dents, and the first time the pattern was drawn, 8 
of these dents would be drawn instead of 14, leaving 
8 dents to be drawn after the draft has been drawn 35 
times, thus making both sides of the cloth with 8 dents 
of plain weave next to the selvages. 

The ends may now be found as follows : 

80 ends per patt X 35 patts = 2800 ends. 



82 

2800 + 4 for the 2 dents over + 48 for the 12 
dents, 4 in a dent, for selvages = 2852 total ends 
required. 

When the cloth contains two or more different 
counts of warp yarns, the reed draft would be made out 
as in the last example ; except that the ends in each 
dent that were of different counts would be placed in a 
separate column as follows : 

Dents Ends. 

B.B. M.B. T.B. 

IC u u u 

The ends in each column would be multiplied by the 
number of patterns, adding selvage ends to that column 
that contained the single yarn, or the ends from which 
the selvage ends would be taken, which is generally the 
bottom beam. 

Mill managers often receive cloth samples which 
they are asked to reproduce and give prices on, etc. 
When this is the case, the reed to use, the picks per 
inch, the counts of the warp and filling yarns must be 
obtained from the sample of cloth furnished. A 
method for finding the counts of yarns by comparison 
will be found on later pages of this book. 

To illustrate a method for finding the ends required 
from a cloth sample, reference will be made to the 
cloth shown in Fig. 1, which is a satin stripe with a 
plain ground. 

As this cloth contains a wide plain stripe, the sley 
reed may be found by the use of a pick glass ; then 
having the sley reed, the dents and ends per dent for 
the satin stripe may be found as follows : 

Suppose the plain stripe counts by glass 64 ends per 
inch, or 64 sley ; then use a thin steel rule that is 
graded in lOOths and measure the satin stripe, which, 
for an example, will be supposed to measure 13 



83 



hundredths of an inch. The number of dents for the 
stripe may be found by the following rule : 

To find the number of dents in a given space, 
when the sley reed is known : 

Rule. Divide the sley reed by 2, residl erjuals the 
dents jjer inch as represented in the cloth ; multiply by the 
space the dents occiqnj in the cloth measured in lOOths, 
ecpmls the number of dents required. 

Fig. 1. 



The dents required for the satin stripe as found by 
this rule are : 

64 sley -^ 2 = 32 dents as represented in 100 
hundredths or 1 inch of cloth, and 32 X .13 = 4.16 
or 4 dents required. There are 16 ends in the stripe 
and 16 ends -^ 4 dents = 4 ends per dent. There 
are 11 dents in the plain stripe, so that one complete 
pattern contains 15 dents and 38 ends. 

Dents. Ends. 

11 22 

15 38 



84 

If the cloth is required to measure 30 inches, the 
ends would be found as follows : 

64 -^ 2 = 32 dents per inch in the cloth, 32 X 30 
inches =960 total dents required. 960 — 12 for 
selvages == 948 dents for body of cloth. 

948 -f- 15 dents per pattern = 63 patterns and 3 
dents over. These 3 dents may be used as previously 
explained by adding them to the 11 dents of plain 
weave in the pattern and using half or 7 dents the first 
time the pattern is drawn, leaving 7 dents to be used 
at the finish of the 63 patterns, which gives 7 dents of 
plain on each side next to the selvages. The ends are 
then found by multiplying the ends per pattern by the 
number of patterns, and adding the ends for the dents 
left over and also the selvage ends. 

38 ends per patt. X 63 patts.= 2394+ 6 +48 = 
2448 total ends required. 

When the cloth contains a narrow plain stripe that 
is less than, a quarter of an inch wide, it will be difficult 
to obtain the sley reed by using a pick glass, so the 
following rule is given : 

To find the sley reed when the number of dents 
and the space they occupy in the cloth, measured 
in lOOths, is known : 

Rule. Divide the number of dents measured by the 
space they occupy in the cloth measured in lOOths, result 
equals the dents per inch a^ rey resented iri the cloth ; 
multiply by 2, equals the sley reed. 

Example. If the space in a cloth that requires 6 
dents measures 15 hundredths of an inch, w^hat is the 
sley of the reed ? 

6 ^ .15 = 40. 
40 X 2 = 80 sley reed. 



s^ 



It must not he understood thnt tlie dents per inch 
in the cloth and dents per incli in the reed mean the 
same thing, for they do not ; for, as the measurements 
have been taken from a ch»th sample, contraction has 
taken i)lace and the dents per inch in the reed will be 
found by a rule ])reviously explained. 

Fig. 2. 




On some cloths, where the sley is very high, and 
especially if ply yarns are used either for backing or 
decorative pui-poses, it is necessary to put several ends 
in the same dent. Sometimes a ply end is drawn in 
the same dent with single yarn, and if the dents are 
very close the ply ends will break the single yarn, 
making it impractical to weave it, etc. When it is 



86 

desired to make a cloth of this construction, it will be 
necessary to make out a reed draft for one complete 
pattern, the desi.s^ner placing the ends in the dents in 
such order as he thinks will be satisfactory. The 
width of the pattern is measured in lOOths of an inch 
and the sley reed obtained as explained when the dents 
and the space they occupy are known. If the sley 
reed found is one that would be considered too fine or 
too coarse for the cloth, a new reed draft is made, and 
more or less ends put in a dent, thus changing the 
number of dents per pattern. The calculation is now 
made again to see if the new draft has brought the sley 
reed to what is considered a more practical reed to 
use on that cloth. 

This idea will be better understood by referring to 
Fig. 2, which shows both sides of a bedford cord cloth 
sample. 

A bedford cord, on its face, appears to be complete 
in one rib, when in fact it requires two, as will be 
understood if the under side or back is examined. 

The sample. Fig. 2, contains 10 face ends and 2 ply 
ends in each rib, and to illustrate the method for find- 
ing the sley reed, two different reed drafts are given. 

Dents. Ends. 

C 1 4 

1 rib. ^1 2 2 ply ends for wadding. 

1 rib. } 

6 20 4 

Supposing two ribs as per draft, or one complete 
pattern, measure 22 hundredths of an inch, the sley reed 
may be found as per rule. 
6 X 100 =: 600. 
600 -^ 22 = 27.27 X 2 = 54.54 or 54 sley reed. 





4 






2 


2p 




4 






4 






2 


2 




4 





87 



Tliis reed may be considered as being too coarse or 
open, so another reed draft is given with less ends per 
dent and two more dents per pattern. 



Ends. 



r 

1 rib. I 
I 



1 ril). -{ 





3 






2 


1 




2 


1 




3 






3 






2 


1 




2 


I 




3 




8. 


20 


4 


Fig. 3. 








8 X 100 =:: 800. 

800 -^ 22 = 36.36 X 2 = 72.72 or 72 sley reed. 

A 72 sley reed would be the best to use as it permits 
of the ply ends being drawn in separate dents. 

To further illustrate this method for finding the ends 
required from a cloth sample, reference will be made 
to Fig. 3, a cloth tiiat combines plain, satin, and leno 
weaves. 

This cloth has a plain section that is wide enough to 
permit the sley reed to be found with a pick glass, but 



88 

for practice in finding the sley reed by measurements, 
the reed will be found by the latter method. 

The plain sections of the cloth contain 1 1 dents and 
22 ends each, and, assuming that each measures 27 
hundredths of an inch 
11 X 100 = 1100. 

1100 -f- 27 = 40.74. 

40.74 X 2 = 81.48 or 82 sley reed. 

Having found the sley reed, it will be necessary to find 
the dents and ends required for the satin and leno stripes. 

Assuming that the satin stripe measures 15 hund- 
redths of an inch, the dents required are 

82 4- 2 = 41. 

41 X .15 = 6.15 or 6 dents, and, as there are 30 
ends in the stripe, 30 -^ 6 = 5 ends per dent. 

Assuming that the leno stripe measures 27 hund- 
redths of an inch 

82 -f- 2 = 41. 

41 X .27= 11.07 or 11 dents. 

In this stripe are 3 dents that contain 2 double ends 
in each dent, therefore there will be 8 empty dents, which, 
divided into 2 spaces, allow 4 dents in each space. 

Starting the pattern at a plain stripe, the reed draft 
is as follows : 

Dents. Ends. 



11 

6 

11 

1 


22 

30 

22 

4 


4 empty 
1 


4 


4 empty 
1 


4 



39 86 

39 dents and 86 ends in one pattern. 



89 

If the cloth were required to he made 27 inches 
wide, the ends would be found as follows : 

82 -^ 2 = 41 dents per inch in the cloth. 

41 X 27 = 1107 — 12 for selvages equals 1095 
dents required for the body of the cloth. 

1095 -^ 39 dents per pattern = 28 patterns and 3 
dents over. These dents may be used by adding them 
to the 1 1 dents of plain weave and drawing 7 dents the 
first time the pattern is drawn, leaving 7 dents to be 
drawn after the pattern has been drawn 28 times. 

86 ends per patt. X 28 = 2408 ends. 

2408 + 6 for the 3 dents + 48 for selvages = 
2462 total ends required. 

Most leno cloths require two or more beams, either 
because the leno ends are of different counts than tlie 
ends in the other weaves in the cloth, or because the 
rate of contraction is not the same. In the cloth, Fig. 
3, the leno ends are the same counts as the ends in the 
plain and satin weaves, and as the leno ends in this 
cloth cross but once in six picks, they will not contract 
any more than the plain weave. If the ends in the 
satin stripe were drawn two in a dent the contraction 
would be less on the stripe than on the plain cloth, but 
to obtain the warp satin effect it is necessary to crowd 
more ends in a dent, thus increasing the sley in the 
stripe. This crowding of the ends increases the .con- 
traction, making it possible to weave such clotlis from 
one beam. 

In the reproduction of cloth from samples it is some- 
times required that the cloth be reproduced on a 
different sley reed than that on which it was woven. 
The object may be to reproduce the cloth in a lighter 
weight without changing the yarns, or it may be to 
make it heavier ; in both cases, however, it is required 
that the width of the pattern l)e the same as is in the 
cloth sample. 



90 

When it is required to reproduce a cloth on a differ- 
ent sley reed, and at the same time, preserve the same 
size of pattern and general appearance of the cloth, it 
will be necessary to measure each section or stripe in 
the pattern and obtain the number of dents that would 
be required in the new reed in each of the different 
parts of the pattern. 

For an illustration. Fig. 1, which was made in a 64 
sley reed and contains 1 1 dents plain and 4 dents satin 
weave will be reproduced on an 80 sley reed. 

Assuming that the plain stripe measures 34 hund- 
redths of an inch, the dents required in an 80 sley 
reed would be 

80 -^ 2 = 40. 

40 X .34 = 13.60 or 14 dents and 28 ends for the 
plain stripe ; then taking the measurement of the satin 
stripe as 13 hundredths of an inch, there would be 

80 -^ 2 r= 40. 

40 X .13 = 5.20 or 5 dents 4 ends in a dent, or 20 
ends for the satin stripe ] one complete pattern would 
contain 

Dents. Ends. 

14 28 

5 20 

19 48 

A dimity cloth is one that has cords that run warp 
way in the cloth, and are formed by drawing several 
ends in one eye of the harness. These ends are run 
on a separate beam, on account of the difference in the 
contraction on them and other ends in the cloth. The 
pattern may be formed by having the cords drawn in 
regular order across the cloth or by having them drawn 
in the reed in groups, etc. 

Sometimes patterns of this kind are required that 
must count a given average sley in the cloth when 



91 

woven, and in that case, it is necessary to find the sley 
reed from the reed draft and the average sley required. 

To find the sley reed to use on a dimity pattern 
when the number of dents and ends in one pattern 
and the average sley cloth required is known : 

Rule. Multiply the average sley by the dents i)cr 
■pattern and divide the result by the ends per pattern; 
multiply this result by two, ecpials the sley reed required. 

Examples. 

If a dimity cloth that contains 10 dents and 23 ends 
per pattern is required to be 116 sley when woven, 
what sley reed is required ? 

^^^ ^ ^^ = 50.43 X 2 = 100.86 or 100 sley reed. 
23 

Find the sley reed and number of ends required to 
make a cloth 30 inches wide and 112 average sley, 
with the folio wins: reed draft. 



;nts. 


Ends. 
B.B. T.B 


6 


12 




3 




2 




3 




2 




3 



11 16 9 

^^^ ^ ^^ = 49.28 X 2 = 98.56 or 98 sley reed. 
25 

Now find ends by rule previously given. 
98 -f- 2 = 49 dents per inch as represented in the 
cloth. 

49 X 30^^ — 1470 total dents requihed. 



92 

1470 — 12 for selvages = 1458 dents for the body 
of the cloth. 

1458 -f- 11 dents per patt. = 132 patts. and 6 dents 
over, which may be used after the pattern has been 
drawn 132 times. 

16 ends per patt. X 132 patts. = 2112 -f 12 for 
the 6 dents + 48 for selvages = 2172 ends in bottom 
beam. 

9 ends per patt. X 132 patts. = 1188 ends in top 
beam. 

2172 + 1188 r=: 3360 total ends. 

3360 -f- 30^^ cloth = 112 average sley. 

The cords in a dimity cloth have a tendency to pre- 
vent contraction in width, and to obtain the correct 
width of the cloth about 8 dents less than the calcu- 
lated number should be used. In the last example, 
1470 — 8 =: 1462 dents would be required. 

COMPARISON OF YARNS FROM 
CLOTH SAMPLES. 

When a cloth has to be reproduced from a small 
sample, the counts of the yarns in the cloth may be 
found by comparison with yarn of which the counts are 
known. 

It is the practice to pull out 6 or 8 ends from the 
cloth and to estimate the counts as near as possible ; 
then to take the same number of ends from a bobbin 
or cop which is of the same counts estimated for the 
ends from the cloth and pass one set of ends around 
the other, twisting each group of ends to form a cord 
of 12 or 16 ends. If the cords appear to be of equal 
thickness it is fair to assume that the counts of the yarn 
from tlie cloth and those from the bobbin or cop are 
the same. If the cord formed bv the ends of the known 



93 

counts is smaller tliai) the cord formed of those taken 
from the cloth, try a coarser counts, if it is larger, try 
a finer counts, until both cords appear to be the same 
size. 

In check patterns, where the fillinij; stripe is required 
to be the same width and have the same appearance as 
to weight, etc., of the warp stripe, it is necessary to put 
in more picks in tlie filling stripe than there are ends 
in the warp stripe, on account of the filling being finer 
than the warp. 

To find the number of picks required in a filling 
stripe to equal the warp stripe in weight : 

Rule. MultlpUj the number of ends in the warp stripe 
by the filling cmmts, and divide the result by the warp 
counts, result equals the number of picks reipiired in the 
filling stripe. 

Example. If a warp satin stripe contains 64 ends 

of 40s warp, how many picks of 60s filling will be 

required to weave a filling satin stripe to equal the 

warp stripe ? 

64 X 60 

=z 96 picks of 60s. 

40 ^ 

PROOF. 

7000 X 64 

— — — z=z 13.33 grains weight of the 64 ends of 

840 X 40 40s warp. 

7000 X 96 

-— — — - = 13.33 grains weight of the 96 picks of 

840 X 60 (50, i^iiing. 

To find the average counts of filling when two or 
more different counts are used : 

Rule. Divide the number of picks of each counts by 
its own counts, add the results, and divide into the total 
jncks per pattern. 



94 

Example. If a cloth contains 38 picks of 15s and 
360 picks of 60s filling in a pattern, what are the 
average counts of the filling ? 

38 -^ 15 = 2.53 relative weight of 15s. 
360 -^ 60 =: 6.00 ^' '^ " 60s. 

398 8^ 

398 4- 8.53 = 46.65 average counts of filling. 

It is the rule in mills, whenever it is possible, to pay 
the operatives by the piece. Weavers are paid by the 
cut, as a general rule, although in mills engaged in 
making very heavy cloths they are sometimes paid by 
the pound. 

A cut may contain almost any length, although 50 
yards is about the average. 

The price paid for weaving will vary with the length 
of the cut, picks per inch, counts of the yarns, speed ol 
the loom, and number of looms it is possible for a 
weaver to run. The speed and number of looms will 
be governed by the counts of the yarns and width of 
the cloth, or, if it is a fancy cloth, the number of 
harnesses and difficulty of the pattern must be taken 
into consideration. Loss in time for mending threads 
and for repairs must also be allowed for, which amount 
will vary with the character of the cloth. 

On plain cloth, the loss in time will vary from 8 to 
20 per cent., while on fancies, the loss may be as high 
as 35 per cent. 

The weekly rate or amount of wages it is desired 
that the weaver should earn is' then fixed, and the price 
per cut may be calculated as follows : 

First find the production of one loom in yards per 
week, multiply by the looms per set, result equals total 
yards ; divide by yards per cut, equals total cuts per 
week. Now divide the weekly rate by the cuts per 
week, equals the weaving price per cut. 



95 

To find the production of a loom : 

Rule. Mult'qybj the speed of the loom by the minutes 
per hour, result equals incks per hour ; multiply by hours 
per week, eipiah jyicks yer week ; multiply by jjer cent, of 
production, eijuals picks per iveek with loss deducted. 
Divide by picks per inch, equals inches per tveek ; divide 
by SQ, equals yards per week ; divide by yards per cut, 
equals cuts ivoven. 

Examples. 

How many 50 yard cuts will a loom weave in 58 
hours with a loss in time of 15 per cent., the loom 
running 160 picks per minute on cloth containing 64 
picks per inch ? 

160 X 60 X 58 X 85% ,,r,, , 

. ^^ = 4.10 cuts per loom per 

64 X 36 X 50 ^eek. 

Find the weaving price per cut from the following 
particulars : 

76 picks per inch. 
1 80 picks per minute. 
85 per cent production. 

8 looms per set. 
50 yards per cut. 
58 hours per week. 
$10.00 weekly rate. 
180 X 60 X 58 X 8 X 85% ^ ^^^^ ^^^^^ ^^^, ^^^^ 
76 X 36 X 50 on 8 looms. 

$10.00 ^ 31.13 = 32.12c. weaving price per cut. 

COST OF PRODUCTION. 

A large portion of the product of cotton mills is 
subject to a variation in the selling price which is fixed 
by the condition of the market. The prices on other 



96 

cloths are fixed by contracts, i. e., agreements to make 
cloths with given constructions at stated prices per 
yard. 

In both instances it is necessary to know the exact 
cost of producing a yard of cloth. In the first case 
this is necessary so that the margin between the 
market price and the actual cost may be known, while 
in the second case it is necessary to know what price 
to l)id on a new cloth when a contract is made, 
Contracts are sometimes made on standard goods, but 
on the finer grades of plain and fancies it is the rule. 

It is the custom in every cotton mill to keep an ac- 
count of the cost of production in every department ; 
the cost is obtained by dividing the production by the 
amount ot the pay roll and is called the labor cost. 

The cost of certain kinds of supplies can very readily 
be charged to tlie departments to which they belong, 
but it may be difficult to apportion a proper share of 
the general expense of running the mill to any one 
department, so for convenience the cost of supplies and 
general expense for the whole mill are added together 
and charged to the weaving room, — the whole expense 
divided by the number of looms, equals the cost per 
loom for supplies and general expense. 

This amount will vary in different mills and will 
average about $1.35 per loom per week on plain to 
$1.80 on fancies. 

Weavers, as a general rule, are paid by the piece, 
but the overseer, second hand, loom fixers, filling men, 
etc., are paid by the day; the amount thus paid is 
divided by the number of looms in the mill and the 
result is the amount paid for oversight, and varies from 
about 30c. to 36c. on plain and 40c. to 46c. per loom 
per week on fancies. 



97 

The cost of the cotton and the lal)Oi- cost on the yarns 
being known, the cost per lb. of a cloth may be found by 
addinjx the additional cost of weavinu;, oversight, and 
general expen.^c. The cost per yard may be found by di- 
viding the cost per lb. l)y the yards of the cloth in one lb. 
The above method for finding cost will be illustrated 
by an example as follows : 

Find the cost of production of a cloth from the 
following particulars. 
84 X 1)2. 

32 inch cloth, 34 ipxhes at reed ) 
60s warp 4% contraction \ estimated. 

80s filling. 
6 looms per set. 
160 picks per minute speed. 
85% production. 
$11.00 weekly rate of wages. 
.16 cotton. 

.07 labor cost on warp. 
.08 " '^ " filling. 
$1.50 general expense per loom per week. 
.40 oversight '' " " i.' 

58 hours per week. 

84 X 32 + 48 = 2736 ends. 
2736 X 104 

———--—- = 5.64 lbs. of warp in 100 yards of 
o^u X bu Q^^^Yi. 

^4c" X 92 X 100 

l^Af^Zro^ == ^-^^ ^^'^- «f filing in 100 yards 

840 X 80 of cloth. 

5.64 + 4.65 = 10.29 lbs. weight of yarns in 100 
yds. of cloth. 

100 yards -^ 10.29 lbs. = 9.71 yards per lb. 

5.64 lbs. of warp -f- 10.29 lbs. total weight = b^% 
warp. 

4.65 lbs. of filling -^ 10.29 lbs. total weight == 45% 
filling. 



98 

16c. cotton X 55%=.0880 cost per lo. of stock for warp 

16c. " x45%=.0720 '' " " " " '' filling 

.1600 total cost per lb. of stock. 

7c. labor cost on warpX 55% =.0385 labor cost per 

lb. on warp. 

8c. labor cost on fillingX 45% =.0360 labor cost per 

lb. on fillins:. 



.0745 total labor cost 
on stock. 

160 X 60 X 58 X 85% ^ .^ .. , ^ , 

— ^-^ = 142.89 yards on 1 loom 

92 X 36 pgj. ^veek. 

142.89 X 6 = 857.34 yards on 6 looms. 

111.00 ^ 857.34 = .0128 weaving cost per yard. 
40c. -^ 142.89 = .0028 oversight " - 

.0156 total cost of weaving 
and oversight per yard. 

.0156 X 9.71 yards per lb. = .1514 cost of weav- 
ing and oversight per lb. 

$1.50 ^ 142.88 yards per week on 1 loom = .0105 
cost of general expense per yard. 

.0105 X 9.71 yards per lb. = .1019 cost of general 
expense per lb. 

Summary of Costs : 
Stock .1600 

Labor cost on stock .0745 
Weavipg and oversight .1514 
General expense '.1019 

.4878 cost per lb. 
.4878 cost per lb. -^ 9.71 yards per lb. = .0502 
cost per yard. 

48.78c. per lb. 
5.02c. " yard. 



99 

EXAMPLES FOR PRACTICE. 

Find the ends required for a cloth 56 sley, 36 inches 
wide, allowing 36 extra ends for selvages. 

2016 ends. 

Find the dents per inch required in a reed for a 72 
sley cloth, with the ends drawn 2 in a dent. 

33.72 dents. 

Find the dents per inch required in a reed to weave 
a cloth 114 sley, with the ends drawn 4 in a dent. 

26.84 dents. 

If a reed contains 1077 dents and is 36 inches long 
inside the reed bars, what sley cloth will it weave with 
the ends drawn 2 in a dent ? 64 sley. 

If a cloth is 76 sley and 30 inches wide when woven, 
with 48 ends in 12 dents, extra for selvages, how wide 
will it be in the reed ? 32.34 inches. 

Find the hanks of filling in 100 yards of cloth woven 
68 X 72, — 30 inches wide. 277.97 hanks. 

Find the lianks of warp in 100 yards of cloth woven 
48 sley, 30 inches wide, 48 ends extra for selvages, and 
7 per cent, contraction. 189.54 hanks. 

If a cloth contains 3000 ends of 60s warp, which 
contract 6 per cent, in weaving, what weight of yarn is 
required to weave 100 yards of cloth ? 6.31 lbs. 

How many lbs. of 36s filling is required to weave 
100 yards of cloth, 36 inches wide in the reed, and 60 
picks per inch? 7.14 lbs. 

If a cloth contains 2000 ends of 60s warp, with 6 per 
cent, contraction and 7 per cent, sizing, 250 ends of 
2 / 20s warp, with 12 per cent, contraction, what weight 
of warp yarn will be required to weave 50 yards of 
cloth? 3.90 lbs. 

l.ofC. 



100 

If a cloth is woven 68 X 72 — 45s warp — 60s 
filling, find the average number by short method. 

51.66 average number. 

If a cloth is woven 36 inches wide at the reed and 
contains 68 picks per inch of 45s fiUincr, and the 
following ends, what is the average number ? 

1800 ends of 40s with 5% contraction. 
250 " ''2/35s " 12% " 

340 " ^•3/15s " 20% " 

30.75 average number. 

Find the average number on a cloth woven 64 X 64 

— 28 inches wide — 7 yards per lb. (Use short 
method.) 33.44 average number. 

Find the average number in the previous example by 
long method, and allow 7 per cent, for contraction on 
warp and 6 per cent, for size, 30 ends extra for selvages 
and 12 dents. 33.25 average number. 

Find the yards per lb. from the following particu- 
lars : (By long method.) 

76 X 80 — 30 inches wide — 60s warp — 95s filling 

— 5% warp contraction and 5% size. 

12.01 yards. 

Find the yards per lb. in the previous example. (By 
short method.) 12.09 yards. 

Find the yards per lb. from the following particu- 
lars : 

1466 ends of 40s with 6% contraction and 5% size. 
100 " "2/20s " 23% 
100 '' ^'2/20s '^ 15%" 
30 inches wide at reed with 68 picks of 50s filling. 

7.98 yards. 
If a piece of cloth, 3x3 inches, weighs 9 grains, 
find the yards per lb. if woven 30 inches wide. 

6.48 yards. 



101 

If the cloth in the previous example was woven 
56 X 60, find the average number. (By short method.) 

30.06 average number. 

If a yard of cloth weighs 896 grains, how many 
yards will weigh 1 lb.? 7.81 yards. 

If a quarter yard of cloth weighs 250 grains, how 
many yards will weigh 1 lb. ? 7 yards. 

Find the counts of fillinsr required to weave a cloth 
to weigh 11 yards per 11)., constructed 76 X 84, 30 
inches wide, 60s warp witli ^\% warp contraction and 
6% size. (By long method.)" 81.05 filling. 

If a cloth is woven 30 inches wide in the reed, with 
76 picks per inch, and requires the following ends, 
what counts of filling is required to make the cloth 
weigh 6i yards per lb. ? 

1950 ends of 50s with 5% contraction. 
300 ^' - 2/20s '• 12% 
200 " '^2 /40s '' 25% ^• 

57.26 filling. 

If a piece of cloth, 4x4 inches, weighs 20 grains 
and counts 64 X 72, make the cloth 30 inches wide, 
using a warp 4 numbers coarser than the average 
numbers, and find the counts of filling required. (Use 
short methods.) ' 33.18 filling. 

Find the weaving price per cut from the following 
particulars : 

Speed of loom 180 picks per minute, 72 picks per 
inch, 7 looms per set, 90% production, 50 yards per 
cut, 58 hour week and $10.00 weekly rate. 

32.84c. per cut. 

Find the per cent, of warp and filling in a cloth 
woven 72 X 78, 60s warp and 100s filling. 

60% warp. 40% filling. 



102 

Find the per cent, of warp and filling in a cloth 
woven 30 inches wide at the reed, with 76 picks of 
70s filling and the following number of ends : 

1850 ends of 60s with 5% contraction. 
250 - ^^ 2/ 15s '' 15% 
300 " -^ 2 /40s " 25% '^ 

73% warp. 27% filling. 



MISCELLANEOUS. 

To find the speed of a loom : 

Rule. MuUiphj the speed of the driving shaft by the 
diameter of the driving pulley ^ and divide the result by the 
diameter of the loom pulley. 

Example. If a loom has a 14 inch pulley and is 
driven by an 8 inch pulley on the driving shaft which 
makes 280 revolutions per minute, what is the speed of 
the loom ? 

280 X 8 

= 160 picks per minute. 

14 

To find the size of loom pulley : 

Rule. Multiply the speed of driving shaft by the 
diameter of driving pulley, a?id divide the result by the 
speed required. 

Example. If the speed of the driving shaft is 280 
R. P. M. and the diameter of the driving pulley is 8 



103 

inches, what size loom pulley is required to liave the 
loom make 160 picks per minute? 

280 X 8 

^14 inch pulley required. 

To find the size of driving pulley : 

Rule. Multiply the speed required by the diameter of 
loom pulley and divide the result by the speed, of the 
drivinor shaft. 

Example. What size driving^ pulley is required on 
the driving shaft which makes 280 R. P. M. to drive a 
loom 160 picks per minute, the loom having a 14 inch 
pulley ? 

:= 8 inch pulley. 

280 

To find the production of a loom : 

Rule. Multiply the speed of the loom by the number 
of minutes it has run, divide the result by picks per inch, 
and this result by 36 inches per yard. 

Example. How many yards will a loom weave in 
10 hours, running 160 picks per minute, on a cloth that 
contains 60 picks per inch ? 

160 X 10 X 60 ^^ ^^ ^ 

= 44.44 yards. 

60 X 36 *" 

It is necessary to make an allowance for stoppages 
on a loom for loss in time, in piecing up ends, changing 
shuttles, etc. If the loom in the last example were 
stopped 10 per cent, of the time, the actual production 
would be as follows : 

160X10X60X.90 ^^ ^ 

=z 40 yards. 

60 X 36 ^ 



104 

When the production per week is desired, the method 
may be shortened by cancellation. The hours and 
minutes per week are placed on the top line and the 
inches per yard under it, and, as they will be present 
in every calculation, they may be cancelled as follows. 
(Mass. has a 58 hour week.) 

96.66 
^12^ = 96.66. 

So that the method with a 58 hour would be 

Speed of loomX 96.66 X % production 

^ =yds. per week. 

picks per inch. 

For a 60 hour week the cancellation would be 

100 

0X11 = 100. 

Use 100 in place of 96.66 as in the case of a 58 
hour week. 

To find the number of teeth required in the pick 
gear on a loom : 

The simplest way is to find the constant, i. e., the 
number of teeth required in the pick gear to put in one 
pick per inch ; then use the constant to find the teeth 
required for any other number of picks. 

To find the constant : 

Rule. Multiphj all the driven gears togetlter and 
divide the result by the product of the drivers and cir- 
aimference of sand roll in indies. Answer equals the 
constant. Leave the pick gear (and carrier gears if there 
are any) out of the calculation. 



105 

When the pick gear is a driver, the constant will be 
a dividend ; when it is a driven gear, tlie constant will 
be a divisor. 

When the constant is a dividend, divide the constant 
by the picks required, and answer equals teeth required 
in pick gear. 

When the constant is a divisor, divide the picks per 
inch required by the constant, and answer equals the 
teeth required in the pick gear. 

Ratchet gears are always driven gears and the sand 
roll is always a driver. 

When the take-up motion is driven from tiie bottom 
shaft, multiply the product of the driven gears by two 
before dividing by the product of the drivers. 

To set the harnesses (on a plain loom) for time and 
position : 

Turn the loom crank to bottom center, loosen the 
cams and turn theni until the harness treadles are 
level ; then turn the loom crank to top center and 
make adjustments on the harness straps to bring the 
shed, that is down, to a point almost touching the race 
board ; turn the loom over and adjust the other slied 
the same way. 

When cover is required, set the harnesses a little 
earlier, i. e., have them level with tlie loom crank a 
little back of tlie bottom center, also raise the wliip-i'oll 
so that it wili be a little above the level of the harness 
eyes when they are level. 

On twills or sateens, have the loom crank on the 
bottom center, then loosen the cams and turn tliem 
until any two of the treadles that are changing become 
level ; make adjustments to obtain the correct position 
of the shed as directed for a plain loom. 



106 

To set the picking motion : 

Have the loom crank on the top center, loosen the 
picking cam and bring it up against the picking ball or 
cone, and make it fast, turn the loom until the picking 
stick is brought forward, adjust the lug straps to bring 
the stick within one and a half inches of the bunter. 
(This distance will vary on different makes of looms.) 
Try the loom and get more or less pick by moving the . 
lug strap up or down on the stick, taking up or letting 
out the lug straps, or moving the picking cam in or 
out on the shaft. 

To set the protector rod : 

Bring the lay forward until the dagger strikes the 
bunter, set the fingers on the protector rod against the 
shuttle box binders and make them fast; adjust the 
spring on the protector rod to give the tension on the 
binders ; put the shuttle in the box and adjust the 
binder (or box front) until the shuttle opens the binder 
sufficient to have the dagger clear the bunter a quarter 
of an inch ; set both sides the same way and set the 
check on the protector rod to hold the shuttle firmly 
in the box. 

To set the filling stop motion : 

Turn the loom to the front center and have the 
shuttle on the filling fork side of the loom ; loosen the 
cam that operates the stop motion finger ; and set it to 
raise the lever and move the finger forward, then turn 
the loom until the lever rests on the smallest diameter 
of the cam; adjust the finger so that there will be a 
quarter of an inch between the end of the finger and 
the hook on the fork, when the lay is forward see that 
the tines of the fork do not touch the grate bars. 

To time the harnesses on a dobby : 
Have the loom crank on the bottom center, loosen 
the dobby crank and place it on either back or front 



107 

center; adjust the connecting rod until the rocker 
arms are vertical ; turn the loom until the bottom knile 
is at its extreme inward position and set the knife a 
quarter of an inch back of the hooks; turn the loom 
over and set the top knife in the same manner. If the 
throw of the knives were changed to obtain either a 
larirer or smaller shed, they must be readjusted to bring 
them a quarter of an inch l)ack of the hooks. 

To time the pattern chain cylinder on a dobby, 
with a worm drive : 

Turn loom until bottom knife is at its extreme 
inward position ; loosen the cylinder and turn until 
the pegs lower the hooks on the knife ; set the worm 
with the straight part of the tooth passing through the 
cylinder gear ; then tighten all the parts. 

To time the pattern chain cylinder on a dobby 
with a pawl and ratchet drive, when the pawl pulls 
the cylinder : 

Have the bottom knife at its extreme inward po- 
sition, loosen the cylinder and turn until the pegs lower 
the hooks on the knife ; loosen the ratchet gear and 
place it in contact with the pawl ; set the stop or check 
wheel to hold cylinder stationary while the pawl 
reaches for another tooth ; tighten all the parts. 

When the pawl reaches for a new tooth, have throw 
enough on the pawl to pass a quarter of an inch 
beyond the tooth so that the top knife will catch tlie 
hooks before the cylinder is moved. 

On box looms, the boxes should commence to move 
when the loom crank is at the bottom center and the 
shuttles are at the box end of the loom, and should 
become stationary when the loom crank reaches the 
top center where the loom commences to pick. 



INDEX. 



Yarn Calculations o 

To find the counts of a cotton yarn when the length and 
weight are known 6 

Reelin(; Yarn 7 

To tind the cotton counts from irregular or short lengths 

of yarn 8 

To find the length in yards when the counts and weight 

are known 10 

To find short lengths when counts and weight in grains 

are known 11 

To find weight when length and counts are known .... 12 

To find the weight of short lengths of cotton yarn . . . lo 

Spun Silk I'-i 

Raw Silk 15 

To find the dram silk counts when length and weight in 

lbs. are known 17 

To find length when counts and weight are known .... 17 
To find the weight when the length and counts are known IS 
To find counts when length and weight in ounces are 

known li> 

To find length when counts and weight in ounces are 

known 1-^ 

To find weight in ounces when length and counts are 

known ^0 

To find the dram silk counts when length and weight in 

grains are known 20 

To find length when counts and weight in grains are 

known -0 

To find weight in grains when length and counts are known 21 
To find the denier counts of raw silk, when length and 

weight in lbs. are known 22 

To find length when weight in lbs. and counts are known . 22 
To find weight when length and counts are known .... 28 
To find the equivalent counts of denier silk in dram silk 

counts 24 



To find the equivalent counts of dram silk in cotton counts 24 
To find the equivalent counts of a denier silk in cotton 

counts 25 

Worsted Yarn 25 

To find the worsted counts from irregular or short lengths 26 

Woolen Yarn 27 

To find counts, length, or weight, apply rules given for 
cotton and worsted yarns (except those for short lengths) 

and use the standard length in the system desired ... 28 

Woolen Cut Yarn 28 

To find the cut counts from short lengths 29 

Woolen Run Yarn 29 

To find counts from short lengths 29 

To find the counts when the weight is known in ounces . 30 
To find the weight in ounces when the length and run 

counts are known 30 

To find the length when counts and weight in ounces are 

known 31 

Linen Yarn 31 

To find the turns of twist per inch 31 

To find the standard breaking weight of warp yarn . ... 32 

Equivalent Counts 33 

To find the equivalent counts of a yarn in any other system 

(except raw silk) 33 

Ply Yarns 34 

Resultant Counts 34 

To find the resultant counts of a ply thread when two or 

more threads of different counts are twisted together . 35 
To find the cotton counts of yarn required to twist with a 
known count of single yarn, to produce a given count of 

ply yarn 36 

To find the weight of each counts of yarn to obtain a given 
weight of ply yarn, when the threads twisted are of un- 
equal counts 37 

Cost of Ply Yarns 39 

To find the cost per lb. of a ply yarn composed of threads 

of different counts and values 39 

Examples for Practice 42 

Cloth Calculations 47 

To find the number of ends required in a warp when the 

sley and width are given 48 

To find the dents per inch for any sley cloth 50 



To find the sley cloth a reed will weave when length and 

total dents in the reed are known 'A 

To find the width of the warp ends in the reed ')■] 

To find the weight of filling in a given length of cloth . . 54 

To find the weight of filling of each color in a checked 
gingham when the picks of each color in the pattern, 
picks per inch, width at reed, and counts of filling are 
known 55 

To find the weight of filling required for a given length of 
cloth when two different counts are used 5() 

Warp Contraction 57 

To find the approximate per cent, of contraction from warp 
to cloth 58 

To find the per cent, to allow for contraction from a cloth 
sample 51) 

To find weight of warp yarn in a given length of cloth . . (iO 

Average Numbers (51 

To find average numbers when sley, picks, width, and 
yards per lb. are known (VI 

To find the average numbers when sley, picks, and counts 
of warp and filling are known (58 

To find average numbers when the cloth contains more 
than one counts of warp, and the number of ends of each 
counts is known, with picks per inch, filling counts, and 
width of cloth at reed (54 

To find the average number when the number of ends and 
picks per inch, width at reed, and yards per lb. are known (55 

To find average number on plain cloths when sley, picks, 

width and yards per lb. are known <57 

To find the average counts of warp yarns when two or 

more different counts are used (i8 

To find the yards per lb., when sley, picks, width and 

counts of warp and filling are known . . . • (58 

To find the yards per lb. when the sley, picks, width, and 

counts of warp and filling are known 70 

To find the yards per lb. from short lengths of cloth ... 71 
To find the counts of filling required to govern the yards 

per lb., when sley, picks, counts of warp, width and yards 

per lb. are known 72 

To find the filling required to govern the yards per lb. (on 

plain cloth) when the sley, picks, width and yards per lb. 

are known 74 

To find the approximate counts of warp and filling from a 

small sample of ordinary plain cloth 75 



To find the per cent, of warp and filling in a piece of cloth, 
is to find the weights of the different yarns used in its 
construction and divide the weight of each by the weight 
of the whole 70 

To find per cent, of warp and filling when ends, picks, and 
counts of warp and filling are known 77 

To find the per cent, of warp and filling in a cloth when 
sley, picks, average counts, and warp counts are known . 78 

To find the per cent, of warp and filling when the cloth 
contains more than one counts of warp, and the counts 
of warp and filling and width at reed are known .... 78 

To find the number of ends in a cloth with an irregular 
reed draft, when the ends in one pattern, the sley reed, 
and the width of the cloth is given 80 

To find the number of dents in a given space, when the 
sley reed is known 88 

To find the sley reed when the number of dents "and the 
space they occupy in the cloth, measured in lOOths, is 
known 84 

To find the sley reed to use on a dimity pattern w4ien the 

number of dents and ends in one pattern and the average 

sley cloth required is known 91 

Comparison of Yarns from Cloth Samples .... 92 
To find the number of picks required in a filling stripe to 

equal the warp stripe in vyeight 93 

To find the average counts of filling when two or more 

different counts are used . 93 

To find the production of a loom 95 

Cost of Production 95 

Examples for Practice 99 

Miscellaneous 102 

To find the speed of a loom 102 

To find the size of loom pulley 102 

To find the size of driving pulley 103 

To find the production of a loom 103 

To find the number of teeth required in the pick gear on a 

loom 104 

To find the constant 104 

To set the harnesses (on a plain loom) for time and position 105 

To set the picking motion 106 

To set the protector rod . . . . • 106 

To set the filling stop motion 106 

To time the harnesses on a dobby 106 

To time the pattern chain cylinder on a dobby, with a 

worm drive .'.... 107 

To time the pattern chain cylinder on a dobby with a pawl 

and ratchet drive, when the pawl pulls the cylinder . . 107 



SfP 7 1904 



